#include <bits/stdc++.h>
using namespace std;
const int maxn = 1500;
int len, anssize = 0;
vector<pair<int, int> > ans;
struct cmp {
  bool operator()(const pair<int, int> &a, const pair<int, int> &b) const {
    return a.second > b.second;
  }
};
long long val[maxn + 1], prefix[maxn + 1], sum, big = 0;
unordered_map<long long,
              priority_queue<pair<int, int>, vector<pair<int, int> >, cmp> >
    all;
int main() {
  all.reserve(10000000);
  cin >> len;
  for (int i = 0; i < len; ++i) cin >> val[i];
  for (int i = 1; i <= len; ++i) {
    prefix[i] = val[i - 1] + prefix[i - 1];
  }
  for (int i = 1; i <= len; ++i) {
    for (int l = 1; l <= i; ++l) {
      all[prefix[i] - prefix[l - 1]].push({l, i});
    }
  }
  for (auto &i : all) {
    vector<pair<int, int> > tmp;
    while (i.second.size()) {
      if (tmp.size() and tmp.back().second >= i.second.top().first)
        i.second.pop();
      else {
        pair<int, int> temp = i.second.top();
        i.second.pop();
        tmp.push_back(temp);
      }
    }
    if (tmp.size() > anssize) {
      ans = tmp;
      anssize = tmp.size();
    }
  }
  cout << anssize << '\n';
  for (int i = 0; i < anssize; ++i) {
    cout << ans[i].first << ' ' << ans[i].second << '\n';
  }
}

### Prompt

Please formulate a CPP solution to the following 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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1500;
int len, anssize = 0;
vector<pair<int, int> > ans;
struct cmp {
  bool operator()(const pair<int, int> &a, const pair<int, int> &b) const {
    return a.second > b.second;
  }
};
long long val[maxn + 1], prefix[maxn + 1], sum, big = 0;
unordered_map<long long,
              priority_queue<pair<int, int>, vector<pair<int, int> >, cmp> >
    all;
int main() {
  all.reserve(10000000);
  cin >> len;
  for (int i = 0; i < len; ++i) cin >> val[i];
  for (int i = 1; i <= len; ++i) {
    prefix[i] = val[i - 1] + prefix[i - 1];
  }
  for (int i = 1; i <= len; ++i) {
    for (int l = 1; l <= i; ++l) {
      all[prefix[i] - prefix[l - 1]].push({l, i});
    }
  }
  for (auto &i : all) {
    vector<pair<int, int> > tmp;
    while (i.second.size()) {
      if (tmp.size() and tmp.back().second >= i.second.top().first)
        i.second.pop();
      else {
        pair<int, int> temp = i.second.top();
        i.second.pop();
        tmp.push_back(temp);
      }
    }
    if (tmp.size() > anssize) {
      ans = tmp;
      anssize = tmp.size();
    }
  }
  cout << anssize << '\n';
  for (int i = 0; i < anssize; ++i) {
    cout << ans[i].first << ' ' << ans[i].second << '\n';
  }
}

```