Problem: This problem is given in two editions, which differ exclusively in the constraints on the number n.

You are given an array of integers a[1], a[2], ..., a[n]. A block is a sequence of contiguous (consecutive) elements a[l], a[l+1], ..., a[r] (1 ≤ l ≤ r ≤ n). Thus, a block is defined by a pair of indices (l, r).

Find a set of blocks (l_1, r_1), (l_2, r_2), ..., (l_k, r_k) such that:

  * They do not intersect (i.e. they are disjoint). Formally, for each pair of blocks (l_i, r_i) and (l_j, r_j) where i ≠ j either r_i < l_j or r_j < l_i. 
  * For each block the sum of its elements is the same. Formally, $$$a[l_1]+a[l_1+1]+...+a[r_1]=a[l_2]+a[l_2+1]+...+a[r_2]= ... = a[l_k]+a[l_k+1]+...+a[r_k].$$$ 
  * The number of the blocks in the set is maximum. Formally, there does not exist a set of blocks (l_1', r_1'), (l_2', r_2'), ..., (l_{k'}', r_{k'}') satisfying the above two requirements with k' > k. 

<image> The picture corresponds to the first example. Blue boxes illustrate blocks.

Write a program to find such a set of blocks.

Input

The first line contains integer n (1 ≤ n ≤ 1500) — the length of the given array. The second line contains the sequence of elements a[1], a[2], ..., a[n] (-10^5 ≤ a_i ≤ 10^5).

Output

In the first line print the integer k (1 ≤ k ≤ n). The following k lines should contain blocks, one per line. In each line print a pair of indices l_i, r_i (1 ≤ l_i ≤ r_i ≤ n) — the bounds of the i-th block. You can print blocks in any order. If there are multiple answers, print any of them.

Examples

Input


7
4 1 2 2 1 5 3


Output


3
7 7
2 3
4 5


Input


11
-5 -4 -3 -2 -1 0 1 2 3 4 5


Output


2
3 4
1 1


Input


4
1 1 1 1


Output


4
4 4
1 1
2 2
3 3

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
using namespace std;
bool compare(pair<int, int> a, pair<int, int> b) { return a.second < b.second; }
void solve(int tim) {
  int n;
  scanf("%d", &n);
  vector<int> a(n + 1);
  for (int i{1}; i < n + 1; i++) scanf("%d", &a[i]);
  vector<int> cumm(n + 1, 0);
  for (int i{1}; i < n + 1; i++) cumm[i] = cumm[i - 1] + a[i];
  map<int, vector<pair<int, int> > > segs;
  for (int i{1}; i < n + 1; i++)
    for (int j{i}; j < n + 1; j++)
      segs[cumm[j] - cumm[i - 1]].push_back({i, j});
  for (map<int, vector<pair<int, int> > >::iterator mit = segs.begin();
       mit != segs.end(); mit++)
    sort(mit->second.begin(), mit->second.end(), compare);
  int maxm = INT_MIN;
  map<int, vector<pair<int, int> > >::iterator mnnit;
  for (map<int, vector<pair<int, int> > >::iterator mit = segs.begin();
       mit != segs.end(); mit++) {
    int lm = -1;
    int cnt = 0;
    for (auto x : mit->second)
      if (x.first > lm) {
        cnt++;
        lm = x.second;
      }
    if (cnt > maxm) {
      maxm = cnt;
      mnnit = mit;
    }
  }
  printf("%d\n", maxm);
  int lm = -1;
  for (auto x : mnnit->second)
    if (x.first > lm) {
      printf("%d %d\n", x.first, x.second);
      lm = x.second;
    }
}
int main() {
  int test = 1;
  for (int t{1}; t < test + 1; t++) {
    solve(t);
  }
}