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:
//package codeforces_464_div2;

import java.util.*;
import java.util.stream.Collectors;

public class F {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int[] arr = new int[n];
        for (int i = 0; i < n; i++) arr[i] = sc.nextInt();
        int[][] sub = new int[n][n];
        for (int i = 0; i < n; i++) {
            sub[i][i] = arr[i];
            for (int j = i + 1; j < n; j++) {
                sub[i][j] = sub[i][j - 1] + arr[j];
            }
        }

        HashMap<Integer, List<P>> hm = new HashMap<>();
        /*for (int i = 0; i < n; i++) {
            for (int j = i; j < n; j++) {
                if (hm.containsKey(sub[i][j])) {
                    hm.get(sub[i][j]).add(new P(i, j));
                } else {
                    List<P> temp = new ArrayList<>();
                    temp.add(new P(i, j));
                    hm.put(sub[i][j], temp);
                }
            }
        }*/
        for(int stop=0; stop<n; stop++) {
            for(int start=0; start<=stop; start++) {
                if (hm.containsKey(sub[start][stop])) {
                    hm.get(sub[start][stop]).add(new P(start, stop));
                } else {
                    List<P> temp = new ArrayList<>();
                    temp.add(new P(start, stop));
                    hm.put(sub[start][stop], temp);
                }
            }
        }

        int ans = Integer.MIN_VALUE;
        /*for(Map.Entry it : hm.entrySet()) {
            int or = overlap(it.getValue());
            ans = Math.max(ans, or);
        }*/

        List<P> ansList = null;
        for (List<P> it : hm.values()) {
            int or = overlap(it);
            if(or>ans) {
                ans = or;
                ansList = it;
            }
        }

        List<P> processedList = extractOverlapping(ansList);

        System.out.println(ans);
        for(int i=0; i<processedList.size(); i++) {
            int A = processedList.get(i).a + 1;
            int B = processedList.get(i).b + 1;
            System.out.println(A + " " + B);
        }
    }

    public static int overlap(List<P> listOfPair) {
        /*List<P> sortedList = listOfPair.stream()
                .sorted((pair1, pair2) -> pair1.getB().compareTo(pair2.getB()))
                .collect(Collectors.toList());*/
        List<P> sortedList = listOfPair;

        int cnt = 0;
        int end = sortedList.get(0).b;
        for (int i = 1; i < sortedList.size(); i++) {
            if (sortedList.get(i).a <= end) {
                cnt++;
            } else {
                end = sortedList.get(i).b;
            }
        }

        return sortedList.size() - cnt;
    }

    public static List<P> extractOverlapping(List<P> ansList) {
        List<P> sortedList = ansList.stream()
                .sorted((pair1, pair2) -> pair1.getB().compareTo(pair2.getB()))
                .collect(Collectors.toList());

        List<P> finalList = new ArrayList<>();
        finalList.add(sortedList.get(0));
        int end = sortedList.get(0).b;
        for (int i = 1; i < sortedList.size(); i++) {
            if (sortedList.get(i).a <= end) {
                continue;
            } else {
                finalList.add(sortedList.get(i));
                end = sortedList.get(i).b;
            }
        }

        return finalList;
    }
}

class P implements Comparable<P> {
    Integer a;
    Integer b;

    public P(int a, int b) {
        this.a = a;
        this.b = b;
    }

    public Integer getA() {
        return a;
    }

    public Integer getB() {
        return b;
    }

    @Override
    public int compareTo(P that) {
        return this.b.compareTo(that.b);
    }
}