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;
long long mult(long long a, long long b, long long p = 1000000007) {
  long long ans = a;
  ans *= b;
  return ((ans >= p) ? ans % p : ans);
}
long long add(long long a, long long b, long long p = 1000000007) {
  long long ans = a + b;
  return ((ans >= p) ? ans - p : ans);
}
long long fpow(long long n, long long k, long long p = 1000000007) {
  long long r = 1;
  for (; k; k >>= 1) {
    if (k & 1) r = r * n % p;
    n = n * n % p;
  }
  return r;
}
long long inv(long long a, long long p = 1000000007) {
  return fpow(a, p - 2, p);
}
int n;
map<int, vector<pair<int, int> > > store;
bool comp(const pair<int, int> &a, const pair<int, int> &b) {
  if (a.second != b.second)
    return a.second < b.second;
  else
    return a.first < b.first;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n;
  vector<int> arr(n);
  vector<int> presum(n + 1, 0);
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
    presum[i + 1] = presum[i] + arr[i];
  }
  for (int i = 0; i < n + 1; i++) {
    for (int j = i + 1; j < n + 1; j++) {
      int sums = presum[j] - presum[i];
      auto it = store.find(sums);
      if (it == store.end()) {
        vector<pair<int, int> > temp;
        temp.push_back({i + 1, j});
        store[sums] = temp;
      } else
        it->second.push_back({i + 1, j});
    }
  }
  int maxans = 0, maxind = -1;
  for (auto &e : store) {
    int locmax = 0;
    int curfirst = -1, cursecond = -1;
    sort(e.second.begin(), e.second.end(), comp);
    for (auto &ee : e.second) {
      if (ee.first <= cursecond)
        continue;
      else {
        locmax++;
        curfirst = ee.first;
        cursecond = ee.second;
      }
    }
    if (locmax > maxans) {
      maxans = locmax;
      maxind = e.first;
    }
  }
  cout << maxans << endl;
  auto it = store.find(maxind);
  int curfirst = -1, cursecond = -1;
  for (auto &ee : it->second) {
    if (ee.first <= cursecond)
      continue;
    else {
      curfirst = ee.first;
      cursecond = ee.second;
      cout << curfirst << " " << cursecond << endl;
    }
  }
  return 0;
}