The only difference between easy and hard versions is a number of elements in the array.

You are given an array a consisting of n integers. The value of the i-th element of the array is a_i.

You are also given a set of m segments. The j-th segment is [l_j; r_j], where 1 ≤ l_j ≤ r_j ≤ n.

You can choose some subset of the given set of segments and decrease values on each of the chosen segments by one (independently). For example, if the initial array a = [0, 0, 0, 0, 0] and the given segments are [1; 3] and [2; 4] then you can choose both of them and the array will become b = [-1, -2, -2, -1, 0].

You have to choose some subset of the given segments (each segment can be chosen at most once) in such a way that if you apply this subset of segments to the array a and obtain the array b then the value max_{i=1}^{n}b_i - min_{i=1}^{n}b_i will be maximum possible.

Note that you can choose the empty set.

If there are multiple answers, you can print any.

If you are Python programmer, consider using PyPy instead of Python when you submit your code.

Input

The first line of the input contains two integers n and m (1 ≤ n ≤ 300, 0 ≤ m ≤ 300) — the length of the array a and the number of segments, respectively.

The second line of the input contains n integers a_1, a_2, ..., a_n (-10^6 ≤ a_i ≤ 10^6), where a_i is the value of the i-th element of the array a.

The next m lines are contain two integers each. The j-th of them contains two integers l_j and r_j (1 ≤ l_j ≤ r_j ≤ n), where l_j and r_j are the ends of the j-th segment.

Output

In the first line of the output print one integer d — the maximum possible value max_{i=1}^{n}b_i - min_{i=1}^{n}b_i if b is the array obtained by applying some subset of the given segments to the array a.

In the second line of the output print one integer q (0 ≤ q ≤ m) — the number of segments you apply.

In the third line print q distinct integers c_1, c_2, ..., c_q in any order (1 ≤ c_k ≤ m) — indices of segments you apply to the array a in such a way that the value max_{i=1}^{n}b_i - min_{i=1}^{n}b_i of the obtained array b is maximum possible.

If there are multiple answers, you can print any.

Examples

Input


5 4
2 -2 3 1 2
1 3
4 5
2 5
1 3


Output


6
2
1 4 


Input


5 4
2 -2 3 1 4
3 5
3 4
2 4
2 5


Output


7
2
3 2 


Input


1 0
1000000


Output


0
0

Note

In the first example the obtained array b will be [0, -4, 1, 1, 2] so the answer is 6.

In the second example the obtained array b will be [2, -3, 1, -1, 4] so the answer is 7.

In the third example you cannot do anything so the answer is 0.
import java.io.*;
import java.util.ArrayList;
import java.util.StringTokenizer;

import static java.lang.Math.*;


public class Main implements Runnable {
    public static void main(String[] args) throws IOException, InterruptedException { Thread t=new Thread(null,new Main(),"solutionThread",1<<28);t.start();
//        System.setIn(new FileInputStream("input.txt"));
//        System.setOut(new PrintStream(new FileOutputStream("output.txt")));
    }

    int n, m;
    long[] ar;
    int[] updated;
    Pair<Integer, Integer>[] v;
    SegmentTree tree;
    @Override
    public void run() {
        Scanner scanner = new Scanner(System.in);
        n = scanner.nextInt();
        m = scanner.nextInt();
        ar = new long[n+5];
        v = new Pair[m+5];
        for (int i = 1; i <= n; i++) {
            ar[i] = scanner.nextInt();
        }
        tree = new MinTree(ar);
        tree.init(1,1,n);
        for (int i = 0; i < m; i++) {
            v[i] = new Pair(scanner.nextInt(), scanner.nextInt());
        }
        updated = new int[m+3];
        long mx = Integer.MIN_VALUE, mxi = -1;
        for (int i = 1; i <= n; i++) {
            partialUpdate(i);
            long x = ar[i];
            x = min(x, tree.query(1,1,n,1, max(1, i-1)));
            x = min(x, tree.query(1,1,n,min(i+1, n), n));
            if (ar[i]-x>mx) {
                mx = ar[i]-x;
                mxi = i;
            }
        }
        System.out.printf("%d\n", mx);
        if (mxi==-1) {
            System.out.println(0);
            return;
        }
        ArrayList<Integer> ansv =  new ArrayList<>();
        for (int i = 0; i < m; i++) {
            int x = v[i].first;
            int y = v[i].second;
            if (x>mxi || y<mxi) {
                ansv.add(i+1);
            }
        }
        System.out.printf("%d\n", ansv.size());
        for (Integer u : ansv) {
            System.out.printf("%d ", u);
        }
    }

    private void partialUpdate(int i) {
        for (int j = 0; j < m; j++) {
            int x = v[j].first;
            int y = v[j].second;
            if (x>i || y<i) {
                if (updated[j]==0) {
                    tree.update(1,1,n,x,y,-1);
                    updated[j] = 1;
                }
            }
            else {
                if (updated[j] == 1) {
                    tree.update(1,1,n,x,y,1);
                    updated[j]=0;
                }
            }
        }
    }
}
class MinTree extends SegmentTree {
    public MinTree(long[] ar) {
        super(ar);
    }
    @Override
    void initialization(int node, int xy) {
        marker[node] = 0;
        tree[node] = ar[xy];
    }
    @Override
    void pushDown(int node, int left, int right) {
        long t = marker[node];
        marker[node] = 0;
        marker[left] += t;
        tree[left] += t;
        marker[right] += t;
        tree[right] += t;
    }
    @Override
    long ignoreQuery() {
        return Long.MAX_VALUE;
    }
    @Override
    long returnQuery(int node) {
        return tree[node];
    }
    @Override
    void ignoreUpdate(int node, int begin, int end, int x, int y, int value) {}
    @Override
    void updateValue(int node, int value) {
        marker[node] += value;
        tree[node] += value;
    }
    @Override
    long merge(long leftValue, long rightValue) {
        return min(leftValue, rightValue);
    }
}
class Pair<S, T> {S first;T second;Pair(S first,T second){this.first=first;this.second=second;}Pair() {}}
abstract class SegmentTree {long[] ar;long[] tree;long[] marker;public SegmentTree(long[] ar){this.ar=ar;tree=new long[ar.length*4];marker=new long[ar.length*4];}void init(int node,int x,int y){if(x==y){initialization(node,x);return;}int left=node * 2,right=node*2+1,mid=(x+y)/2;init(left,x,mid);init(right,mid+1,y);tree[node]=merge(tree[left],tree[right]);}void update(int node,int begin,int end,int x,int y,int value){if(x>y){ignoreUpdate(node,begin,end,x,y,value);return;}if(begin==x && end==y){updateValue(node,value);return;}int left=node * 2,right=node*2+1,mid=(begin+end)/2;pushDown(node,left,right);update(left,begin,mid,x,min(mid,y),value);update(right,mid+1,end,max(mid+1,x),y,value);tree[node]=merge(tree[left],tree[right]);}long query(int node,int begin,int end,int x,int y){if(x>y)return ignoreQuery();if(begin==x && end==y)return returnQuery(node);int left=node*2,right=node*2+1,mid=(begin+end)/2;pushDown(node,left,right);long leftValue=query(left,begin,mid,x,min(mid,y));long rightValue=query(right,mid+1,end,max(mid+1,x),y);return merge(leftValue,rightValue);}abstract void initialization(int node,int xy);abstract void pushDown(int node,int left,int right);abstract long ignoreQuery();abstract long returnQuery(int node);abstract void ignoreUpdate(int node,int begin,int end,int x,int y,int value);abstract void updateValue(int node,int value);abstract long merge(long leftValue,long rightValue);}
class Scanner {BufferedReader br=new BufferedReader(new InputStreamReader(System.in));StringTokenizer st=new StringTokenizer("");Scanner(InputStream in){}String next(){while(!st.hasMoreElements())try{st=new StringTokenizer(br.readLine());}catch(IOException e){e.printStackTrace();}return st.nextToken();}int nextInt(){return Integer.parseInt(next());}long nextLong(){return Long.parseLong(next());}double nextDouble(){return Double.parseDouble(next());}}
class System {static InputStream in=java.lang.System.in;static PrintStream out=new PrintStream(java.lang.System.out);static void setIn(FileInputStream i){java.lang.System.setIn(i);}static void setOut(PrintStream p){java.lang.System.setOut(p);}}