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.*;
import java.text.*;
//Solution Credits: Taranpreet Singh
public class Main{
    //SOLUTION BEGIN
    void pre() throws Exception{}
    void solve(int TC) throws Exception{
        int n = ni(), m = ni();
        int[] a = new int[n];
        for(int i = 0; i< n; i++)a[i] = ni();
        LazySegTree t = new LazySegTree(a);
        int[][] seg = new int[m][], segl = new int[m][], segr = new int[m][];
        for(int i = 0; i< m; i++){
            int l = ni()-1, r = ni()-1;
            t.rng(l, r, 0, t.m-1, 1, -1);
            seg[i] = new int[]{l,r};
            segl[i] = new int[]{l,r};
            segr[i] = new int[]{l,r};
        }
        Arrays.sort(segl, (int[] i1, int[] i2) -> Integer.compare(i1[0], i2[0]));
        Arrays.sort(segr, (int[] i1, int[] i2) -> Integer.compare(i1[1], i2[1]));
        long ans= 0;int ind = 0;
        for(int i = 0, j = 0, k = 0; i< n; i++){
            while(j<m && segl[j][0]<=i){
                t.rng(segl[j][0], segl[j][1], 0, t.m-1, 1, 1);
                j++;
            }
            while(k< m && segr[k][1]<i){
                t.rng(segr[k][0], segr[k][1], 0, t.m-1, 1, -1);
                k++;
            }
            long x = t.sum(0, n, 0, t.m-1, 1);
            if(a[i]-x>ans){
                ans = a[i]-x;
                ind = i;
            }
        }
        pn(ans);
        StringBuilder aa = new StringBuilder("");int c = 0;
        for(int i = 0; i< m; i++)if(seg[i][0]>ind || seg[i][1]<ind){
            aa.append((i+1)+" ");c++;
        }
        pn(c+"\n"+aa.toString());
    }
    class LazySegTree{
        int m = 1;
        long[] t, lazy;
        public LazySegTree(int[] a){
            while(m<a.length)m<<=1;
            t = new long[m<<1];
            lazy = new long[m<<1|1];
            for(int i = 0; i< a.length; i++)t[m+i] = a[i];
            for(int i = a.length; i< m; i++)t[m+i] = MX;
            for(int i = m-1; i> 0; i--)t[i] = Math.min(t[i<<1], t[i<<1|1]);
        }
        void rng(int l, int r, int ll, int rr, int i, long x){
            if(lazy[i]!=0){
                t[i] += lazy[i];
                if(ll!=rr){
                    lazy[i<<1] += lazy[i];
                    lazy[i<<1|1] += lazy[i];
                }
                lazy[i] = 0;
            }
            if(r<ll || rr<l)return;
            if(l<=ll && rr<=r){
                t[i] += x;
                if(ll!=rr){
                    lazy[i<<1] += x;
                    lazy[i<<1|1] += x;
                }
                return;
            }
            int mid = (ll+rr)>>1;
            rng(l,r,ll,mid,i<<1,x);rng(l,r,mid+1,rr,i<<1|1,x);
            t[i] = Math.min(t[i<<1],t[i<<1|1]);
        }
        long sum(int l, int r, int ll, int rr, int i){
            if(lazy[i]!=0){
                t[i] += lazy[i];
                if(ll!=rr){
                    lazy[i<<1] += lazy[i];
                    lazy[i<<1|1] += lazy[i];
                }
                lazy[i] = 0;
            }
            if(r<ll || rr<l)return MX;
            if(l<=ll && rr<=r)return t[i];
            int mid = (ll+rr)>>1;
            return Math.min(sum(l,r,ll,mid,i<<1),sum(l,r,mid+1,rr,i<<1|1));
        }
    }
    //SOLUTION END
    void hold(boolean b)throws Exception{if(!b)throw new Exception("Hold right there, Sparky!");}
    long mod = (long)1e9+7, IINF = (long)4e18;
    final int INF = (int)2e9, MX = (int)1e6+1;
    DecimalFormat df = new DecimalFormat("0.000000000000000");
    double PI = 3.1415926535897932384626433832792884197169399375105820974944, eps = 1e-8;
    static boolean multipleTC = false, memory = false;
    FastReader in;PrintWriter out;
    void run() throws Exception{
        in = new FastReader();
        out = new PrintWriter(System.out);
        int T = (multipleTC)?ni():1;
        //Solution Credits: Taranpreet Singh
        pre();for(int i = 1; i<= T; i++)solve(i);
        out.flush();
        out.close();
    }
    public static void main(String[] args) throws Exception{
        if(memory)new Thread(null, new Runnable() {public void run(){try{new Main().run();}catch(Exception e){e.printStackTrace();}}}, "1", 1 << 28).start();
        else new Main().run();
    }
    long gcd(long a, long b){return (b==0)?a:gcd(b,a%b);}
    int gcd(int a, int b){return (b==0)?a:gcd(b,a%b);}
    int bit(long n){return (n==0)?0:(1+bit(n&(n-1)));}
    void p(Object o){out.print(o);}
    void pn(Object o){out.println(o);}
    void pni(Object o){out.println(o);out.flush();}
    String n()throws Exception{return in.next();}
    String nln()throws Exception{return in.nextLine();}
    int ni()throws Exception{return Integer.parseInt(in.next());}
    long nl()throws Exception{return Long.parseLong(in.next());}
    double nd()throws Exception{return Double.parseDouble(in.next());}

    class FastReader{
        BufferedReader br;
        StringTokenizer st;
        public FastReader(){
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        public FastReader(String s) throws Exception{
            br = new BufferedReader(new FileReader(s));
        }

        String next() throws Exception{
            while (st == null || !st.hasMoreElements()){
                try{
                    st = new StringTokenizer(br.readLine());
                }catch (IOException  e){
                    throw new Exception(e.toString());
                }
            }
            return st.nextToken();
        }

        String nextLine() throws Exception{
            String str = "";
            try{   
                str = br.readLine();
            }catch (IOException e){
                throw new Exception(e.toString());
            }  
            return str;
        }
    }
}