// Original C++ solution (slower):
//#include<bits/stdc++.h>

#include<cstdio>

using namespace std;

typedef long long ll;

#define rep(i, n) for(int i = 0; i < (n); i++)

#define rep1(i, n) for(int i = 1; i <= (n); i++)

#define co(x) cout << (x) << "\n"

#define cosp(x) cout << (x) << " "

#define ce(x) cerr << (x) << "\n"

#define cesp(x) cerr << (x) << " "

#define pb push_back

#define mp make_pair

#define Would

#define you

#define please



const int cm = 1 << 17;

char cn[cm], * ci = cn + cm, ct;

inline char getcha() {

	if (ci - cn == cm) { fread(cn, 1, cm, stdin); ci = cn; }

	return *ci++;

}

inline int getint() {

	int A = 0;

	if (ci - cn + 16 > cm) while ((ct = getcha()) >= '0') A = A * 10 + ct - '0';

	else while ((ct = *ci++) >= '0') A = A * 10 + ct - '0';

	return A;

}



void pakuripair_sort(int N, int A[], int B[]) {

	const int b = 8;

	int tmp[100001], tmp2[100001];

	rep(k, 4) {

		int kazu[1 << b] = {}, kazu2[1 << b] = {};

		rep(i, N) kazu[A[i] >> k * b & ((1 << b) - 1)]++;

		rep(i, (1 << b) - 1) kazu[i + 1] += kazu[i];

		for (int i = N - 1; i >= 0; i--) {

			tmp[--kazu[A[i] >> k * b & ((1 << b) - 1)]] = A[i];

			tmp2[kazu[A[i] >> k * b & ((1 << b) - 1)]] = B[i];

		}

		k++;

		rep(i, N) kazu2[tmp[i] >> k * b & ((1 << b) - 1)]++;

		rep(i, (1 << b) - 1) kazu2[i + 1] += kazu2[i];

		for (int i = N - 1; i >= 0; i--) {

			A[--kazu2[tmp[i] >> k * b & ((1 << b) - 1)]] = tmp[i];

			B[kazu2[tmp[i] >> k * b & ((1 << b) - 1)]] = tmp2[i];

		}

	}

}



int main() {

	//cin.tie(0);

	//ios::sync_with_stdio(false);





	int N = getint(), M = getint();

	int A[100001], AN[100001];

	A[0] = 0;

	rep1(i, N) {

		int a = getint();

		A[i] = (A[i - 1] + a) % M;

		AN[i] = i;

	}

	pakuripair_sort(N, A + 1, AN + 1);



	int B[100001];

	int mae = A[0];

	int k = 0;

	ll kotae = 0;

	rep(i, N + 1) {

		if (mae != A[i]) k++;

		B[AN[i]] = k;

		mae = A[i];

	}

	int C[100001] = {};

	rep(i, N + 1) {

		kotae += C[B[i]]++;

	}

	printf("%lld", kotae);



	Would you please return 0;

}

// Optimized solution to the same problem (faster):
//#include<bits/stdc++.h>

#include<cstdio>

using namespace std;

typedef long long ll;

#define rep(i, n) for(int i = 0; i < (n); i++)

#define rep1(i, n) for(int i = 1; i <= (n); i++)

#define co(x) cout << (x) << "\n"

#define cosp(x) cout << (x) << " "

#define ce(x) cerr << (x) << "\n"

#define cesp(x) cerr << (x) << " "

#define pb push_back

#define mp make_pair

#define Would

#define you

#define please



const int cm = 1 << 17;

char cn[cm], * ci = cn + cm, ct;

inline char getcha() {

	if (ci - cn == cm) { fread(cn, 1, cm, stdin); ci = cn; }

	return *ci++;

}

inline int getint() {

	int A = 0;

	if (ci - cn + 16 > cm) while ((ct = getcha()) >= '0') A = A * 10 + ct - '0';

	else while ((ct = *ci++) >= '0') A = A * 10 + ct - '0';

	return A;

}



void pakuri_sort(int N, int A[]) {

	const int b = 8;

	int tmp[200001];

	rep(k, 4) {

		int kazu[1 << b] = {}, kazu2[1 << b] = {};

		rep(i, N) kazu[A[i] >> k * b & ((1 << b) - 1)]++;

		rep(i, (1 << b) - 1) kazu[i + 1] += kazu[i];

		for (int i = N - 1; i >= 0; i--) tmp[--kazu[A[i] >> k * b & ((1 << b) - 1)]] = A[i];

		k++;

		rep(i, N) kazu2[tmp[i] >> k * b & ((1 << b) - 1)]++;

		rep(i, (1 << b) - 1) kazu2[i + 1] += kazu2[i];

		for (int i = N - 1; i >= 0; i--) A[--kazu2[tmp[i] >> k * b & ((1 << b) - 1)]] = tmp[i];

	}

}



int main() {

	//cin.tie(0);

	//ios::sync_with_stdio(false);





	int N = getint(), M = getint();

	int A[100001];

	A[0] = 0;

	rep1(i, N) {

		int a = getint();

		A[i] = (A[i - 1] + a) % M;

	}

	pakuri_sort(N, A + 1);



	//数だけでええんか

	int mae = -1;

	int k = 0;

	ll kotae = 0;

	rep(i, N + 1) {

		if (mae != A[i]) k = 0;

		kotae += k++;

		mae = A[i];

	}

	printf("%lld", kotae);



	Would you please return 0;

}

// Fastest known solution for this problem: