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 int maxn = 1505;
struct node {
  int l, r;
  node(int l, int r) : l(l), r(r) {}
  friend bool operator<(node a, node b) {
    return a.r == b.r ? a.l < b.l : a.r < b.r;
  }
};
long long sum[maxn];
map<long long, int> vis;
vector<int> V;
vector<node> Q[2250005];
vector<node> ans[2250005];
int cnt = 0;
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int t;
    scanf("%d", &t);
    sum[i] = sum[i - 1] + t;
  }
  for (int i = 1; i <= n; i++) {
    for (int j = i; j <= n; j++) {
      long long t = sum[j] - sum[i - 1];
      if (!vis[t]) vis[t] = ++cnt;
      Q[vis[t]].push_back(node(i, j));
    }
  }
  int maxx = 0, p;
  for (int i = 1; i <= cnt; i++) {
    sort(Q[i].begin(), Q[i].end());
    int last = 0;
    for (auto t : Q[i]) {
      if (t.l > last) {
        last = t.r;
        ans[i].push_back(t);
      }
    }
    if (ans[i].size() > maxx) {
      maxx = ans[i].size();
      p = i;
    }
  }
  cout << ans[p].size() << endl;
  for (auto t : ans[p]) {
    cout << t.l << " " << t.r << endl;
  }
  return 0;
}