3 6 5 3 2 8 3
Yes 3 3 1 4 No Yes 2 0 1 For the first test case, if we set $A=(3,1,4)$, then * $k=0$: Setting $(i,j)=(1,1)$, we have $0\equiv 3+3\pmod 6$. * $k=1$: Setting $(i,j)=(1,3)$, we have $1\equiv 3+4\pmod 6$. * $k=2$: Setting $(i,j)=(3,3)$, we have $2\equiv 4+4\pmod 6$. * $k=3$: There is no pair of indices $(i,j)$ that satisfies the condition. * $k=4$: Setting $(i,j)=(1,2)$, we have $4\equiv 3+1\pmod 6$. * $k=5$: Setting $(i,j)=(2,3)$, we have $5\equiv 1+4\pmod 6$. Thus, there are exactly $5$ values of $0\le k < 6$ that satisfy the condition.
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