Maksim has nn objects and mm boxes, each box has size exactly kk. Objects are numbered from 11 to nn in order from left to right, the size of the ii-th object is aiai.

Maksim wants to pack his objects into the boxes and he will pack objects by the following algorithm: he takes one of the empty boxes he has, goes from left to right through the objects, and if the ii-th object fits in the current box (the remaining size of the box is greater than or equal to aiai), he puts it in the box, and the remaining size of the box decreases by aiai. Otherwise he takes the new empty box and continues the process above. If he has no empty boxes and there is at least one object not in some box then Maksim cannot pack the chosen set of objects.

Maksim wants to know the maximum number of objects he can pack by the algorithm above. To reach this target, he will throw out the leftmost object from the set until the remaining set of objects can be packed in boxes he has. Your task is to say the maximum number of objects Maksim can pack in boxes he has.

Each time when Maksim tries to pack the objects into the boxes, he will make empty all the boxes he has before do it (and the relative order of the remaining set of objects will not change).

The first line of the input contains three integers nn, mm, kk (1≤n,m≤2⋅1051≤n,m≤2⋅105, 1≤k≤1091≤k≤109) — the number of objects, the number of boxes and the size of each box.

The second line of the input contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤k1≤ai≤k), where aiai is the size of the ii-th object.

Print the maximum number of objects Maksim can pack using the algorithm described in the problem statement.

In the first example Maksim can pack only 44 objects. Firstly, he tries to pack all the 55 objects. Distribution of objects will be [5],[2,1][5],[2,1]. Maxim cannot pack the next object in the second box and he has no more empty boxes at all. Next he will throw out the first object and the objects distribution will be [2,1],[4,2][2,1],[4,2]. So the answer is 44.

In the second example it is obvious that Maksim cannot pack all the objects starting from first, second, third and fourth (in all these cases the distribution of objects is [4][4]), but he can pack the last object ([1][1]).

In the third example Maksim can pack all the objects he has. The distribution will be [1,2],[3],[1,1][1,2],[3],[1,1].

代码：

#include <bits/stdc++.h>

using namespace std;

const int maxn = 2e5 + 10;

int N, M, K;

int a[maxn];

int main() {

    scanf("%d%d%d", &N, &M, &K);

    for(int i = 1; i <= N; i ++)

        scanf("%d", &a[i]);

    int cnt = 0;

    int ans = K;

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

        if(a[i] <= ans) {

            ans -= a[i];

            cnt ++;

        } else {

            M --;

            if(!M) break;

            ans = K - a[i];

            cnt ++;

        }

    }

    printf("%d\n", cnt);

    return 0;

}