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 long long Mod = 1000000007;
const double eps = 1e-9;
inline int sign(double a) { return a < -eps ? -1 : a > eps; }
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long ex_gcd(long long a, long long b, long long &x, long long &y) {
  if (b > a) return ex_gcd(b, a, y, x);
  if (b == 0) {
    x = 1;
    y = 0;
    return a;
  } else {
    long long x1, y1;
    long long res = ex_gcd(b, a % b, x1, y1);
    x = y1;
    y = x1 - a / b * y1;
    return res;
  }
}
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
long long ksc(long long x, long long y) {
  long long res = 0;
  while (y) {
    if (y & 1) res = (res + x) % Mod;
    x = (x + x) % Mod;
    y >>= 1;
  }
  return res % Mod;
}
long long Pow(long long n, long long x) {
  long long res = 1;
  while (x) {
    if (x & 1) res = ksc(res, n);
    n = ksc(n, n);
    x >>= 1;
  }
  return res;
}
int num[2000];
long long psum[2000];
map<long long, vector<pair<int, int>>> res;
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", num + i);
  }
  for (int i = 1; i <= n; i++) psum[i] = psum[i - 1] + num[i];
  for (int i = 1; i <= n; i++) {
    for (int j = i; j >= 1; j--) {
      if (res[psum[i] - psum[j - 1]].empty() ||
          res[psum[i] - psum[j - 1]].back().second < j)
        res[psum[i] - psum[j - 1]].push_back(make_pair(j, i));
    }
  }
  map<long long, vector<pair<int, int>>>::iterator mit = res.begin();
  map<long long, vector<pair<int, int>>>::iterator res_it = res.begin();
  for (; mit != res.end(); mit++) {
    if (mit->second.size() > res_it->second.size()) res_it = mit;
  }
  printf("%d\n", res_it->second.size());
  int s = 0;
  for (int i = 0; i < res_it->second.size(); i++) {
    printf("%d %d\n", res_it->second[i].first, res_it->second[i].second);
  }
  return 0;
}