Problem: 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.

Solution (in JAVA):

Solution:
import java.util.*;
import java.io.*;

public class ArraysAndSegmentsHard {
    public static void main(String[] args) {
        PrintWriter out = new PrintWriter(System.out);
        FastScanner scanner = new FastScanner();
        int N = scanner.nextInt();
        int M = scanner.nextInt();
        ArrayList<Integer>[] right = new ArrayList[N];
        ArrayList<Integer>[] left = new ArrayList[N];
        SegTree ST = new SegTree(N);
        int[] arr = new int[N];
        for(int i= 0; i < N; i++) {
            arr[i] = scanner.nextInt();
            ST.increment(i,i,arr[i]);
            left[i] = new ArrayList<>();
            right[i] = new ArrayList<>();
        }
        int[][] segments = new int[M][2];
        for(int i = 0; i < M; i++) {
            int l = scanner.nextInt()-1;
            int r = scanner.nextInt()-1;
            segments[i][0] = l;
            segments[i][1] = r;
            left[l].add(i);
            right[r].add(i);
        }
        long ans = 0;
        int ind = -1;
        for(int i=0;i<N;i++){
            for(int x:left[i])
                ST.increment(segments[x][0], segments[x][1], -1);
            long min=ST.minimum(i, i);
            long max=ST.maximum(0, N-1);
            if(max-min>ans){
                ans=max-min;
                ind=i;
            }
            for(int x:right[i])
                ST.increment(segments[x][0],segments[x][1], 1);
        }
        out.println(ans);
        ArrayList<Integer> outList = new ArrayList<>();
        for(int i = 0; i < M; i++) {
            if (segments[i][0] <= ind && segments[i][1] >= ind) {
                outList.add(i);
            }
        }
        out.println(outList.size());
        for(int x: outList) {
            out.print((x+1) + " ");
        }
        out.println();
        out.flush();
    }
    public static class SegTree {
        int[] hi, lo;
        long[] delta, min, max;
        int size;
        public SegTree(int ss) {
            size = ss;
            hi = new int[4*ss+1];
            lo = new int[4*ss+1];
            min = new long[4*ss+1];
            delta = new long[4*ss+1];
            max = new long[4*ss+1];
            init(1, 0, ss-1);
        }
        public void init(int cur, int l, int r) {
            lo[cur] = l;
            hi[cur] = r;
            //base case
            if (l==r) return;
            //else recurse
            int mid = (l+r)/2;
            init(cur*2, l, mid);
            init(cur*2+1, mid+1, r);
        }
        public void increment(int l, int r, long amt) {
            increment(1, l, r, amt);
        }
        public void increment(int cur, int l, int r, long amt) {
            //total coverage
            if (l > hi[cur] || r < lo[cur]) return;
            //complete coverage
            if (r >= hi[cur] && l <= lo[cur]) {
                delta[cur] += amt;
                return;
            }
            //partial coverage -> recurse
            prop(cur);
            increment(cur*2, l , r, amt);
            increment(cur*2+1, l , r, amt);
            update(cur);
        }
        public long minimum(int l, int r) {
            return minimum(1, l, r);
        }
        public long minimum(int cur, int l, int r) {
            //no coverage
            if (l > hi[cur] || r < lo[cur]) return Long.MAX_VALUE;
            //total coverage
            if (r >= hi[cur] && l <= lo[cur]) return min[cur] + delta[cur];
            //partial coverage
            prop(cur);
            long left = minimum(cur*2, l, r);
            long right = minimum(cur*2+1, l, r);
            update(cur);
            return Math.min(left, right);
        }
        public long maximum(int l, int r) {
            return maximum(1, l, r);
        }
        public long maximum(int cur, int l, int r) {
            //no coverage
            if (l > hi[cur] || r < hi[cur]) return Long.MIN_VALUE;
            //total coverage
            if (r >= lo[cur] && l <= hi[cur]) return max[cur] + delta[cur];
            //partial coverage
            prop(cur);
            long left = maximum(cur*2, l, r);
            long right = maximum(cur*2+1, l, r);
            update(cur);
            return Math.max(left, right);
        }
        public void prop(int cur) {
            delta[cur*2]+=delta[cur];
            delta[cur*2+1] += delta[cur];
            delta[cur] = 0;
        }
        public void update(int cur) {
            min[cur] = Math.min(min[cur*2]+delta[2*cur], min[cur*2+1] + delta[cur*2+1]);
            max[cur] = Math.max(max[cur*2]+delta[2*cur], max[cur*2+1] + delta[cur*2+1]);
        }
    }
    
    public static class FastScanner {
        BufferedReader br;
        StringTokenizer st;
        
        public FastScanner(Reader in) {
            br = new BufferedReader(in);
        }
        
        public FastScanner() {
            this(new InputStreamReader(System.in));
        }
        
        String next() {
            while (st == null || !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());
        }
        
        String readNextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
        
        int[] readIntArray(int n) {
            int[] a = new int[n];
            for (int idx = 0; idx < n; idx++) {
                a[idx] = nextInt();
            }
            return a;
        }
    }
}