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;
const long long N = 200005;
const long long M = 1505;
const long long mod = 1000000007;
const long long inf = (1 << 30);
const long long linf = (1LL << 61);
int n, v[M], x, nr, l[M * M];
vector<pair<long long, long long> > a, ans;
map<int, vector<pair<long long, long long> > > mp;
int main() {
  ios::sync_with_stdio(false);
  cin.tie();
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> v[i], v[i] += v[i - 1];
  for (int j = 1; j <= n; ++j) {
    for (int i = 1; i <= j; ++i) {
      x = v[j] - v[i - 1];
      if (mp[x].size() == 0) l[++nr] = x;
      mp[x].push_back({i, j});
    }
  }
  for (int i = 1; i <= nr; ++i) {
    a.resize(0);
    for (int j = 0; j < mp[l[i]].size(); ++j) {
      if (!a.size() || a[a.size() - 1].second < mp[l[i]][j].first)
        a.push_back(mp[l[i]][j]);
    }
    if (a.size() > ans.size()) ans = a;
  }
  cout << ans.size() << '\n';
  for (int i = 0; i < ans.size(); ++i)
    cout << ans[i].first << ' ' << ans[i].second << '\n';
  return 0;
}