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 CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
inline int read() {
  char c = getchar();
  int s = 0;
  while (c < '0' or c > '9') c = getchar();
  while (c >= '0' and c <= '9') s = s * 10 + c - '0', c = getchar();
  return s;
}
const int N = (1 << 12) + 5;
const int B = 256;
int n, q, a[N];
int pos[N];
int tot, cnt;
int stx[N * B * 2], sty[N * B * 2], top;
int hb(int x, int y) {
  if (!x | !y) return x | y;
  stx[++tot] = x;
  sty[tot] = y;
  return tot;
}
struct node {
  vector<int> id;
  map<int, int> ma[N];
  int cx(int l, int r) {
    if (l > id.back()) return 0;
    if (r < id.front()) return 0;
    int x = lower_bound(id.begin(), id.end(), l) - id.begin();
    int y = upper_bound(id.begin(), id.end(), r) - id.begin() - 1;
    if (x > y) return 0;
    return ma[x][y];
  }
  friend node operator+(node x, node y) {
    node z;
    int head = 0, tou = 0;
    while (head < x.id.size() and tou < y.id.size()) {
      if (x.id[head] < y.id[tou]) {
        z.id.push_back(x.id[head]);
        head++;
      } else {
        z.id.push_back(y.id[tou]);
        tou++;
      }
    }
    while (head < x.id.size()) {
      z.id.push_back(x.id[head]);
      head++;
    }
    while (tou < y.id.size()) {
      z.id.push_back(y.id[tou]);
      tou++;
    }
    for (int i = 0; i < z.id.size(); i++) {
      for (int k = i + 1; k < z.id.size(); k++) {
        z.ma[i][k] = hb(x.cx(z.id[i], z.id[k]), y.cx(z.id[i], z.id[k]));
      }
      z.ma[i][i] = hb(x.cx(z.id[i], z.id[i]), y.cx(z.id[i], z.id[i]));
    }
    return z;
  }
} t[N / B + 1];
int ans[N * B * 2];
int x, y;
node fh(int u) {
  node g;
  g.id.push_back(pos[u]);
  g.ma[0][0] = pos[u];
  return g;
}
node ycl(int l, int r) {
  if (l == r) return fh(l);
  return (ycl(l, ((l + r) >> 1)) + ycl(((l + r) >> 1) + 1, r));
}
int main() {
  n = read();
  q = read();
  for (int i = 1; i <= n; i++) {
    a[i] = read();
    pos[a[i]] = i;
  }
  cnt = n / B + 1;
  tot = n;
  for (int i = 1; i <= cnt; i++) {
    if ((i - 1) * B + 1 > n) {
      cnt--;
      break;
    }
    t[i] = ycl((i - 1) * B + 1, min(n, i * B));
  }
  for (int i = 1; i <= q; i++) {
    x = read();
    y = read();
    for (int k = 1; k <= cnt; k++) {
      ans[i] = hb(ans[i], t[k].cx(x, y));
    }
  }
  cout << tot << endl;
  for (int i = n + 1; i <= tot; i++) {
    printf("%d %d\n", stx[i], sty[i]);
  }
  for (int i = 1; i <= q; i++) {
    printf("%d ", ans[i]);
  }
  return 0;
}