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 N = 5e6;
const long long INF = 1e8, mod = 100000000;
int n, m;
long long a[2600];
struct node {
  long long l, r;
  long long k;
} b[N];
int cnt;
bool cmp(node x, node y) { return x.r < y.r; }
map<long long, long long> mp;
map<long long, long long> mmp;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &a[i]);
  }
  for (int i = 1; i <= n; i++) {
    long long s = 0;
    for (int j = i; j <= n; j++) {
      s += a[j];
      b[++cnt].l = i, b[cnt].r = j;
      b[cnt].k = s;
    }
  }
  sort(b + 1, b + 1 + cnt, cmp);
  long long ans = INF * INF;
  int x = 0;
  for (int i = 1; i <= cnt; i++) {
    if (b[i].l <= mmp[b[i].k]) continue;
    mp[b[i].k]++;
    mmp[b[i].k] = b[i].r;
    int y = mp[b[i].k];
    if (y > x) {
      ans = b[i].k;
      x = y;
    }
  }
  if (x == 0) {
    cout << -1 << endl;
    return 0;
  }
  cout << mp[ans] << endl;
  int num = 0;
  for (int i = 1; i <= cnt; i++) {
    if (b[i].k != ans || b[i].l <= num) continue;
    num = b[i].r;
    printf("%lld %lld\n", b[i].l, b[i].r);
  }
  return 0;
}