#include <iostream>
#include <algorithm>

using namespace std;

typedef long long LL;

const int N = 1e5 + 10;

int n, k;
LL s[N];
int cnt[N];

int main()
{
    scanf("%d%d", &n, &k);
    for (int i = 1; i <= n; i ++ ) 
    {
        scanf("%d", &s[i]);
        s[i] += s[i - 1];               //求前缀和
    }

    LL res = 0;
    cnt[0] = 1 ;                        //默认第一个位置(s[0])，有(0%k==0), 所以cnt[0]=1;
    for (int i = 1; i <= n; i ++ )
    {
        res += cnt[s[i] % k];           //若两个前缀和对k取模后余数相同，则这两个前缀和相减对k取模为0
        cnt[s[i] % k] ++ ;          
    }

    printf("%lld\n", res);

    return 0;
}