_Hands that shed innocent blood!_
There are _n_ guilty people in a line, the _i_\-th of them holds a claw with length _L__i_. The bell rings and every person kills some of people in front of him. All people kill others at the same time. Namely, the _i_\-th person kills the _j_\-th person if and only if _j_ < _i_ and _j_ ≥ _i_ - _L__i_.
You are given lengths of the claws. You need to find the total number of alive people after the bell rings.
## Input
The first line contains one integer _n_ (1 ≤ _n_ ≤ 106) — the number of guilty people.
Second line contains _n_ space-separated integers _L_1, _L_2, ..., _L__n_ (0 ≤ _L__i_ ≤ 109), where _L__i_ is the length of the _i_\-th person's claw.
## Output
Print one integer — the total number of alive people after the bell rings.
[samples]
## Note
In first sample the last person kills everyone in front of him.
_Hands that shed innocent blood!_
有 #cf_span[n] 个有罪的人排成一列，第 #cf_span[i] 个人手持一把长度为 #cf_span[Li] 的爪子。钟声响起，每个人都会杀死他前面的一些人。所有人同时行动。具体来说，第 #cf_span[i] 个人会杀死第 #cf_span[j] 个人，当且仅当 #cf_span[j < i] 且 #cf_span[j ≥ i - Li]。
给你每个人的爪子长度，你需要求出钟声响起后仍然活着的人的总数。
第一行包含一个整数 #cf_span[n] (#cf_span[1 ≤ n ≤ 106]) —— 有罪之人的数量。
第二行包含 #cf_span[n] 个用空格分隔的整数 #cf_span[L1, L2, ..., Ln] (#cf_span[0 ≤ Li ≤ 109])，其中 #cf_span[Li] 是第 #cf_span[i] 个人的爪子长度。
请输出一个整数——钟声响起后仍然活着的人的总数。
在第一个样例中，最后一个人杀死了他前面的所有人。
## Input
第一行包含一个整数 #cf_span[n] (#cf_span[1 ≤ n ≤ 106]) —— 有罪之人的数量。第二行包含 #cf_span[n] 个用空格分隔的整数 #cf_span[L1, L2, ..., Ln] (#cf_span[0 ≤ Li ≤ 109])，其中 #cf_span[Li] 是第 #cf_span[i] 个人的爪子长度。
## Output
请输出一个整数——钟声响起后仍然活着的人的总数。
[samples]
## Note
在第一个样例中，最后一个人杀死了他前面的所有人。
Let $ n $ be the number of people, indexed from $ 1 $ to $ n $.
Let $ L_i $ denote the claw length of the $ i $-th person.
For each person $ i $, they kill all persons $ j $ such that:
$$
j < i \quad \text{and} \quad j \geq i - L_i
$$
That is, person $ i $ kills all people in the range $ [\max(1, i - L_i), i - 1] $.
A person $ j $ is alive after the bell rings if and only if **no** person $ i > j $ kills them, i.e., for all $ i > j $, it is **not** true that $ j \in [i - L_i, i - 1] $.
Equivalently, person $ j $ is killed if there exists some $ i > j $ such that:
$$
i - L_i \leq j < i
$$
Thus, person $ j $ is **alive** if and only if:
$$
\forall i > j, \quad j < i - L_i \quad \text{or} \quad j \geq i
$$
But since $ j < i $ always holds for $ i > j $, the condition simplifies to:
$$
\forall i > j, \quad j < i - L_i
\quad \iff \quad
\forall i > j, \quad L_i < i - j
$$
Alternatively, person $ j $ is **killed** if:
$$
\exists i > j \text{ such that } L_i \geq i - j
$$
So, person $ j $ is alive if and only if:
$$
\max_{i > j} (i - L_i) > j
$$
is **false** — that is, if:
$$
\max_{i > j} (i - L_i) \leq j
$$
But more directly:
Define for each $ i $, the leftmost index that person $ i $ can kill: $ \ell_i = \max(1, i - L_i) $.
Then person $ j $ is killed if there exists $ i > j $ such that $ \ell_i \leq j $.
So, person $ j $ is alive if and only if:
$$
\min_{i > j} \ell_i > j
$$
Thus, the number of alive people is:
$$
\left| \left\{ j \in \{1, 2, \dots, n\} \mid \min_{i > j} \max(1, i - L_i) > j \right\} \right|
$$
Note: For $ j = n $, there is no $ i > j $, so the minimum over an empty set is $ \infty $, so person $ n $ is always alive.
We can compute this efficiently by processing from right to left.
Let $ R_j = \min_{i > j} \max(1, i - L_i) $, with $ R_n = \infty $.
Then:
- Person $ j $ is alive $ \iff R_j > j $
We can compute $ R_j $ iteratively from $ j = n-1 $ down to $ 1 $:
$$
R_j = \min\left( \max(1, j+1 - L_{j+1}), R_{j+1} \right)
$$
**Final formal statement:**
Let $ n \in \mathbb{N} $, $ L_1, \dots, L_n \in \mathbb{Z}_{\geq 0} $.
Define $ \ell_i = \max(1, i - L_i) $ for $ i = 1, \dots, n $.
Define $ R_n = \infty $, and for $ j = n-1, \dots, 1 $:
$$
R_j = \min(\ell_{j+1}, R_{j+1})
$$
Then the number of alive people is:
$$
\sum_{j=1}^n \mathbf{1}_{\{R_j > j\}}
$$