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>
#pragma GCC optimize("unroll-loops")
using namespace std;
void err(istream_iterator<string> it) {}
template <typename Tem1, typename... Args>
void err(istream_iterator<string> it, Tem1 a, Args... args) {
  err(++it, args...);
}
bool big_cmp(int a, int b) { return a > b; }
const long long mod = 1e9 + 7;
const long long inf = 1ll << 60;
const int maxn = 2e5 + 5;
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  int n;
  cin >> n;
  vector<int> a(n);
  for (int i = 0; i < (n); i++) cin >> a[i];
  map<int, vector<pair<int, int> > > tmp;
  for (int i = 0; i < (n); i++) {
    map<int, int> mp;
    int nat = 0;
    for (int j = i; j < n; j++) {
      nat += a[j];
      if (!mp.count(nat)) {
        mp[nat] = j;
      }
    }
    for (auto &ele : mp) {
      tmp[ele.first].push_back({i, ele.second});
    }
  }
  int mxv = 0;
  vector<pair<int, int> > rev;
  for (auto &ele : tmp) {
    vector<pair<int, int> > &v = ele.second;
    sort(v.begin(), v.end(), [](pair<int, int> a, pair<int, int> b) {
      if (a.second != b.second) return a.second < b.second;
      return a.first < b.first;
    });
    int mId = -1;
    int nv = 0;
    vector<pair<int, int> > bb;
    for (int i = 0; i < (v.size()); i++) {
      if (v[i].first > mId) {
        nv++;
        mId = v[i].second;
        bb.push_back({v[i].first, v[i].second});
      }
    }
    if (nv > mxv) {
      mxv = nv;
      rev = bb;
    }
  }
  cout << mxv << '\n';
  for (int i = 0; i < (rev.size()); i++) {
    cout << rev[i].first + 1 << ' ' << rev[i].second + 1 << '\n';
  }
}