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 JAVA):

Solution:
import java.util.*;
public class MyClass {
    static class Pair implements Comparable<Pair>{
        int x,y;
        Pair(int x,int y){
            this.x=x; this.y=y;
        }
        public int compareTo(Pair p){
            return Integer.compare(this.y,p.y);
        }
    }
    public static void main(String args[]) {
      Scanner in =new Scanner(System.in);
      HashMap<Long,ArrayList<Pair>> hm=new HashMap<Long,ArrayList<Pair>>();
      int n=in.nextInt();
      int a[]=new int[n];
      long sum[]=new long[n];
      long s=0;
      for(int i=0;i<n;i++){  a[i]=in.nextInt(); s+=a[i]; sum[i]=s; }
      for(int i=0;i<n;i++){
          for(int j=i;j<n;j++){
              long x=sum[j]-sum[i]+a[i];
              ArrayList<Pair> temp=new ArrayList<Pair>();
              if(hm.containsKey(x)) { temp=hm.get(x); }
              temp.add(new Pair(i+1,j+1));
              hm.put(x,temp);
          }
      }
      ArrayList<Pair> ans=new ArrayList<Pair>();
      for(Map.Entry em:hm.entrySet()){
          ArrayList<Pair> array=hm.get(em.getKey());
          Collections.sort(array);
          int prev=0;
          ArrayList<Pair> temp=new ArrayList<Pair>();
          for(int i=0;i<array.size();i++){
              if(array.get(i).x>prev){ temp.add(new Pair(array.get(i).x,array.get(i).y)); prev=array.get(i).y; }
          }
        //  System.out.println(temp.size());
          if(temp.size()>ans.size()){
              ans=(ArrayList<Pair>)temp.clone();
          }
      }
      long g=-5;
      System.out.println(ans.size()); 
      for(int i=0;i<ans.size();i++){
          System.out.println(ans.get(i).x+" "+ans.get(i).y);
      }
    }
}