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;
int read() {
  int x = 0, sgn = 1;
  char ch = getchar();
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') sgn = -1;
  for (; isdigit(ch); ch = getchar()) x = (x << 1) + (x << 3) + (ch ^ 48);
  return x * sgn;
}
const int N = (1 << 12) + 10;
const int M = (1 << 16) + 10;
const int OP = 2.2e6 + 10;
const int BL = (1 << 8) + 10;
int tot, c[OP], d[OP], q, n, a[N], pos[N], ans[M];
int merge(int x, int y) {
  if (!x || !y) return x + y;
  return c[++tot] = x, d[tot] = y, tot;
}
struct BLOCK {
  vector<vector<int> > id;
  vector<int> val;
  void init(int x) {
    val.resize(x);
    id.resize(x);
    for (int i = 0; i < x; i++) id[i].resize(x - i);
  }
  BLOCK(int x = -1) {
    if (~x) init(1), val[0] = pos[x], id[0][0] = pos[x] + 1;
  }
  int ask(int l, int r) {
    if (l > val.back() || r < val.front()) return 0;
    int x = lower_bound(val.begin(), val.end(), l) - val.begin();
    int y = upper_bound(val.begin(), val.end(), r) - val.begin() - 1;
    if (x > y)
      return 0;
    else
      return id[x][y - x];
  }
  friend BLOCK operator+(BLOCK a, BLOCK b) {
    BLOCK ans;
    ans.init(a.val.size() + b.val.size());
    merge(a.val.begin(), a.val.end(), b.val.begin(), b.val.end(),
          ans.val.begin());
    for (int i = 0; i < ans.val.size(); i++)
      for (int j = i; j < ans.val.size(); j++)
        ans.id[i][j - i] =
            merge(a.ask(ans.val[i], ans.val[j]), b.ask(ans.val[i], ans.val[j]));
    return ans;
  }
} B[N];
BLOCK solve(int l, int r) {
  if (l == r) return BLOCK(l);
  int mid = l + r >> 1;
  return solve(l, mid) + solve(mid + 1, r);
}
int main() {
  n = read();
  q = read();
  int blo = sqrt(q);
  tot = n;
  for (int i = 0; i < n; i++) a[i] = read() - 1, pos[a[i]] = i;
  for (int i = 0; i <= (n - 1) / blo; i++)
    B[i] = solve(i * blo, min(n - 1, (i + 1) * blo - 1));
  for (int i = 0; i < q; i++) {
    int l = read() - 1, r = read() - 1;
    for (int j = 0; j <= (n - 1) / blo; j++)
      ans[i] = merge(ans[i], B[j].ask(l, r));
  }
  printf("%d\n", tot);
  for (int i = n + 1; i <= tot; i++) printf("%d %d\n", c[i], d[i]);
  for (int i = 0; i < q; i++) printf("%d%c", ans[i], " \n"[i == q - 1]);
  return 0;
}