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;
using ll = long long;
using ld = long double;
using vi = vector<int>;
using vll = vector<ll, ll>;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
ll modpow(ll a, ll b, ll mod = (ll)(1e9 + 7)) {
  if (!b) return 1;
  a %= mod;
  return modpow(a * a % mod, b / 2, mod) * (b & 1 ? a : 1) % mod;
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const int DEBUG = 1;
const int mxn = 1501;
int n;
int a[mxn], b[mxn];
map<int, int> source;
map<int, int> dp[mxn];
void solve() {
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < n; i++) {
    int x = 0;
    for (int j = i; j <= n - 1; j++) {
      x += a[j];
      (dp[j + 1][x]) = max((dp[j + 1][x]), (source[x] + 1));
    }
    for (pii pp : dp[i + 1])
      (source[pp.first]) = max((source[pp.first]), (pp.second));
  }
  int x = 0;
  for (pii pp : source)
    if (source[x] < pp.second) x = pp.first;
  cout << source[x] << endl;
  vector<pii> schedule;
  for (int i = 0; i < n; i++) {
    int j = i, y = a[j++];
    while (j < n && y != x) y += a[j++];
    if (y == x) schedule.push_back(pii(i, j));
  }
  auto cmp = [](pii a, pii b) { return a.second < b.second; };
  sort(begin(schedule), end(schedule), cmp);
  int i = 0;
  for (pii pp : schedule) {
    if (i <= pp.first) {
      cout << pp.first + 1 << " " << pp.second << endl;
      i = pp.second;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  while (cin >> n) solve();
  return 0;
}