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 n, A[1505], dp[1505];
vector<long long> vals;
vector<pair<long long, long long> > segs[100005 * 25];
vector<pair<long long, long long> > ans[100005 * 25];
void elim() {}
long long BS(long long u) {
  long long l = 0, r = vals.size() - 1;
  while (r > l) {
    long long mid = (l + r) / 2;
    if (vals[mid] < u)
      l = mid + 1;
    else
      r = mid;
  }
  return l;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (long long i = 1; i <= n; i++) cin >> A[i];
  for (long long i = 1; i <= n; i++) dp[i] = dp[i - 1] + A[i];
  for (long long i = 1; i <= n; i++)
    for (long long j = i; j <= n; j++) vals.push_back(dp[j] - dp[i - 1]);
  sort(vals.begin(), vals.end());
  vector<long long> temp = {vals[0]};
  for (long long i = 1; i < vals.size(); i++)
    if (vals[i] != vals[i - 1]) temp.push_back(vals[i]);
  vals = temp;
  for (long long i = 1; i <= n; i++)
    for (long long j = i; j <= n; j++) {
      long long x = BS(dp[j] - dp[i - 1]);
      segs[x].push_back({j, i});
    }
  for (long long ix = 0; ix < vals.size(); ix++) {
    vector<pair<long long, long long> > V = segs[ix];
    sort(V.begin(), V.end());
    vector<pair<long long, long long> > lis;
    vector<pair<long long, long long> > sav;
    lis.assign(V.size() + 1, {1, -1});
    sav.assign(V.size() + 1, {4444444444444444444, -1});
    sav[0] = {0, -1};
    long long lisans = 0;
    for (long long i = 0; i < V.size(); i++) {
      long long b = V[i].second, e = V[i].first;
      long long l = 0, r = sav.size();
      while (r > l) {
        long long mid = (l + r) / 2;
        if (sav[mid].first < b)
          l = mid + 1;
        else
          r = mid;
      }
      l--;
      lisans = max(lisans, l + 1);
      lis[i] = {l + 1, sav[l].second};
      sav[l + 1] = min(make_pair(e, i), sav[l + 1]);
    }
    long long u = sav[lisans].second;
    while (u > -1) {
      ans[ix].push_back(V[u]);
      u = lis[u].second;
    }
  }
  vector<pair<long long, long long> > finans = {};
  long long ind = -1;
  for (long long i = 0; i < vals.size(); i++)
    if (finans.size() < ans[i].size()) {
      finans = ans[i];
      ind = vals[i];
    }
  cout << finans.size() << "\n";
  for (pair<long long, long long> P : finans)
    cout << P.second << " " << P.first << "\n";
}