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;
unordered_map<int, pair<int, pair<int, int>>> m;
void insertion(int value, int count, pair<int, int> index) {
  if (m.find(value) == m.end())
    m[value] = make_pair(count, index);
  else {
    if (count > m[value].first) m[value] = make_pair(count, index);
  }
}
int main() {
  int n;
  cin >> n;
  vector<int> a(n);
  vector<vector<int>> sum(n), d(n);
  vector<vector<pair<int, int>>> trace(n);
  cin >> a[0];
  sum[0] = vector<int>(n);
  d[0] = vector<int>(n);
  trace[0] = vector<pair<int, int>>(n, make_pair(-1, -1));
  for (int i = 1; i < n; ++i) {
    cin >> a[i];
    a[i] += a[i - 1];
    sum[i] = vector<int>(n);
    d[i] = vector<int>(n);
    trace[i] = vector<pair<int, int>>(n, make_pair(-1, -1));
  }
  for (int j = 0; j < n; ++j) {
    sum[0][j] = a[j];
    d[0][j] = 1;
  }
  for (int i = 1; i < n; ++i) {
    for (int j = i; j < n; ++j) sum[i][j] = a[j] - a[i - 1];
  }
  int max_count = 1;
  pair<int, int> in = make_pair(0, 0);
  for (int i = 1; i < n; ++i) {
    for (int k = 0; k < i; ++k)
      insertion(sum[k][i - 1], d[k][i - 1], make_pair(k, i - 1));
    for (int j = i; j < n; ++j) {
      if (m.find(sum[i][j]) == m.end())
        d[i][j] = 1;
      else {
        d[i][j] = m[sum[i][j]].first + 1;
        trace[i][j] = m[sum[i][j]].second;
      }
      if (d[i][j] > max_count) {
        max_count = d[i][j];
        in = make_pair(i, j);
      }
    }
  }
  cout << max_count << endl;
  for (int i = 0; i < max_count; ++i) {
    cout << in.first + 1 << " " << in.second + 1 << endl;
    in = trace[in.first][in.second];
  }
}