Problem: You are given a permutation a_1, a_2, ..., a_n of numbers from 1 to n. Also, you have n sets S_1,S_2,..., S_n, where S_i=\\{a_i\}. Lastly, you have a variable cnt, representing the current number of sets. Initially, cnt = n.

We define two kinds of functions on sets:

f(S)=min_{u∈ S} u;

g(S)=max_{u∈ S} u.

You can obtain a new set by merging two sets A and B, if they satisfy g(A)<f(B) (Notice that the old sets do not disappear).

Formally, you can perform the following sequence of operations:

  * cnt← cnt+1;

  * S_{cnt}=S_u∪ S_v, you are free to choose u and v for which 1≤ u, v < cnt and which satisfy g(S_u)<f(S_v).




You are required to obtain some specific sets.

There are q requirements, each of which contains two integers l_i,r_i, which means that there must exist a set S_{k_i} (k_i is the ID of the set, you should determine it) which equals \\{a_u∣ l_i≤ u≤ r_i\}, which is, the set consisting of all a_i with indices between l_i and r_i.

In the end you must ensure that cnt≤ 2.2× 10^6. Note that you don't have to minimize cnt. It is guaranteed that a solution under given constraints exists.

Input

The first line contains two integers n,q (1≤ n ≤ 2^{12},1 ≤ q ≤ 2^{16}) — the length of the permutation and the number of needed sets correspondently.

The next line consists of n integers a_1,a_2,⋅⋅⋅, a_n (1≤ a_i≤ n, a_i are pairwise distinct) — given permutation.

i-th of the next q lines contains two integers l_i,r_i (1≤ l_i≤ r_i≤ n), describing a requirement of the i-th set.

Output

It is guaranteed that a solution under given constraints exists.

The first line should contain one integer cnt_E (n≤ cnt_E≤ 2.2× 10^6), representing the number of sets after all operations.

cnt_E-n lines must follow, each line should contain two integers u, v (1≤ u, v≤ cnt', where cnt' is the value of cnt before this operation), meaning that you choose S_u, S_v and perform a merging operation. In an operation, g(S_u)<f(S_v) must be satisfied.

The last line should contain q integers k_1,k_2,⋅⋅⋅,k_q (1≤ k_i≤ cnt_E), representing that set S_{k_i} is the ith required set.

Please notice the large amount of output.

Examples

Input


3 2
1 3 2
2 3
1 3


Output


6
3 2
1 3
5 2
4 6 

Input


2 4
2 1
1 2
1 2
1 2
1 1


Output


5
2 1
2 1
2 1
5 3 3 1

Note

In the first sample:

We have S_1=\{1\},S_2=\{3\},S_3=\{2\} initially.

In the first operation, because g(S_3)=2<f(S_2)=3, we can merge S_3,S_2 into S_4=\{2,3\}.

In the second operation, because g(S_1)=1<f(S_3)=2, we can merge S_1,S_3 into S_5=\{1,2\}.

In the third operation, because g(S_5)=2<f(S_2)=3, we can merge S_5,S_2 into S_6=\{1,2,3\}.

For the first requirement, S_4=\{2,3\}=\\{a_2,a_3\}, satisfies it, thus k_1=4.

For the second requirement, S_6=\{1,2,3\}=\\{a_1,a_2,a_3\}, satisfies it, thus k_2=6

Notice that unused sets, identical sets, outputting the same set multiple times, and using sets that are present initially are all allowed.

Solution (in JAVA):

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

public class H {
    static final boolean RUN_TIMING = false;
    static char[] inputBuffer = new char[1 << 6];
    static PushbackReader in = new PushbackReader(new BufferedReader(new InputStreamReader(System.in)), 1 << 6);
    static PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
    ArrayList<Long> pairs = new ArrayList<>();
    ArrayList<Integer> sets = new ArrayList<>();

    public void go() throws IOException {
        // in = new PushbackReader(new BufferedReader(new FileReader(new File("test.txt"))), 1 << 20);
        // out = new PrintWriter(new FileWriter(new File("output.txt")));
        int n = ipar();
        int q = ipar();
        int[] arr = iapar(n);
        int[] pos = new int[n+1];
        for (int i = 0; i < n; i++) {
            pos[arr[i]] = i+1;
        }
        SegmentTree tree = new SegmentTree(pos, n);
        for (int i = 0; i < q; i++) {
            int l = ipar();
            int r = ipar();
            sets.add(tree.query(l, r));
        }
        out.println(n + pairs.size());
        for (long p : pairs) {
            out.print(p/50000000L);
            out.print(" ");
            out.println(p%50000000L);
        }
        for (int x : sets) {
            out.print(x);
            out.print(" ");
        }
        out.println();
    }

    private class SegmentTree {
        private class Node {
            TreeSet<Integer> indices;
            HashMap<Long, Integer> map;

            public Node() {
                indices = new TreeSet<>();
                map = new HashMap<>();
            }
        }

        Node[] tree;
        int n;
        int currIndex;

        public SegmentTree(int[] pos, int n) {
            this.n = n;
            tree = new Node[n*4];
            currIndex = n+1;
            build(1, 1, n, pos);
        }

        private void build(int v, int tl, int tr, int[] pos) {
            if (tl == tr) {
                tree[v] = new Node();
                tree[v].indices.add(pos[tl]);
                tree[v].map.put(pos[tl]*50000000L + pos[tl], pos[tl]);
                return;
            }
            int tm = (tl+tr)/2;
            build(v*2, tl, tm, pos);
            build(v*2+1, tm+1, tr, pos);
            tree[v] = new Node();
            tree[v].indices.addAll(tree[v*2].indices);
            tree[v].indices.addAll(tree[v*2+1].indices);
        }

        private int query(int v, int tl, int tr, int l, int r) {
            Integer lbObj = tree[v].indices.ceiling(l);
            Integer rbObj = tree[v].indices.floor(r);
            if (lbObj == null || rbObj == null) {
                return 0;
            }
            int lb = lbObj.intValue();
            int rb = rbObj.intValue();
            if (lb > rb) {
                return 0;
            }
            long p = lb*50000000L + rb;
            Integer ans = tree[v].map.get(p);
            if (ans != null) {
                return ans;
            }
            
            int tm = (tl+tr)/2;
            int lSet = query(v*2, tl, tm, l, r);
            int rSet = query(v*2+1, tm+1, tr, l, r);
            if (lSet == 0) {
                tree[v].map.put(p, rSet);
                return rSet;
            }
            if (rSet == 0) {
                tree[v].map.put(p, lSet);
                return lSet;
            }

            tree[v].map.put(p, currIndex);
            pairs.add(lSet*50000000L + rSet);
            return currIndex++;
        }
        
        public int query(int l, int r) {
            return query(1, 1, n, l, r);
        }
    }

    public int ipar() throws IOException {
        return Integer.parseInt(spar());
    }

    public int[] iapar(int n) throws IOException {
        int[] arr = new int[n];
        for (int i = 0; i < n; i++) {
            arr[i] = ipar();
        }
        return arr;
    }

    public long lpar() throws IOException {
        return Long.parseLong(spar());
    }

    public long[] lapar(int n) throws IOException {
        long[] arr = new long[n];
        for (int i = 0; i < n; i++) {
            arr[i] = lpar();
        }
        return arr;
    }

    public double dpar() throws IOException {
        return Double.parseDouble(spar());
    }

    public String spar() throws IOException {
        int len = 0;
        int c;
        do {
            c = in.read();
        } while (Character.isWhitespace(c) && c != -1);
        if (c == -1) {
            throw new NoSuchElementException("Reached EOF");
        }
        do {
            inputBuffer[len] = (char)c;
            len++;
            c = in.read();
        } while (!Character.isWhitespace(c) && c != -1);
        while (c != '\n' && Character.isWhitespace(c) && c != -1) {
            c = in.read();
        }
        if (c != -1 && c != '\n') {
            in.unread(c);
        }
        return new String(inputBuffer, 0, len);
    }

    public String linepar() throws IOException {
        int len = 0;
        int c;
        while ((c = in.read()) != '\n' && c != -1) {
            if (c == '\r') {
                continue;
            }
            inputBuffer[len] = (char)c;
            len++;
        }
        return new String(inputBuffer, 0, len);
    }

    public boolean haspar() throws IOException {
        String line = linepar();
        if (line.isEmpty()) {
            return false;
        }
        in.unread('\n');
        in.unread(line.toCharArray());
        return true;
    }

    public static void main(String[] args) throws IOException {
        long time = 0;
        time -= System.nanoTime();
        new H().go();
        time += System.nanoTime();
        if (RUN_TIMING) {
            System.out.printf("%.3f ms%n", time / 1000000.0);
        }
        out.flush();
        in.close();
    }
}