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;
const int MAXN = 5005;
const int MAXQ = 2e5 + 5;
const int Limit = 2.2e6;
template <typename T>
void chkmax(T &x, T y) {
  x = max(x, y);
}
template <typename T>
void chkmin(T &x, T y) {
  x = min(x, y);
}
template <typename T>
void read(T &x) {
  x = 0;
  int f = 1;
  char c = getchar();
  for (; !isdigit(c); c = getchar())
    if (c == '-') f = -f;
  for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
  x *= f;
}
struct info {
  int l[MAXN], r[MAXN];
  vector<vector<int>> home;
  void init(int x) {
    home.resize(x + 1);
    for (int i = 0; i <= x; i++) home[i].resize(x + 1);
  }
  int query(int ql, int qr) {
    if (l[ql] > r[qr])
      return 0;
    else
      return home[l[ql]][r[qr]];
  }
} res[MAXN];
int n, q, size, a[MAXN], b[MAXN], k[MAXQ];
int tot, block, ql[MAXN], qr[MAXN], belong[MAXN];
vector<pair<int, int>> ans;
int merge(int x, int y) {
  if (x == 0 || y == 0) return x + y;
  ans.emplace_back(x, y);
  assert(++size <= Limit);
  return size;
}
info getres(int l, int r) {
  if (l == r) {
    info ans;
    for (int i = 1; i <= n; i++) {
      ans.l[i] = 1 + (i > b[l]);
      ans.r[i] = (i >= b[r]);
    }
    ans.init(1);
    ans.home[1][1] = b[l];
    return ans;
  }
  int mid = (l + r) / 2;
  info ans, ansl, ansr;
  ansl = getres(l, mid);
  ansr = getres(mid + 1, r);
  for (int i = 1, j = 1, k = 0; i <= n; i++) {
    k += (l <= a[i] && a[i] <= r);
    ans.l[i] = j, ans.r[i] = k;
    j += (l <= a[i] && a[i] <= r);
  }
  ans.init(r - l + 1);
  for (int i = 1; i <= n; i++)
    if (l <= a[i] && a[i] <= r) {
      for (int j = i; j <= n; j++)
        if (l <= a[j] && a[j] <= r) {
          ans.home[ans.l[i]][ans.r[j]] =
              merge(ansl.query(i, j), ansr.query(i, j));
        }
    }
  return ans;
}
int main() {
  read(n), read(q), size = n;
  for (int i = 1; i <= n; i++) read(a[i]), b[a[i]] = i;
  block = 1 << 8;
  for (int i = 1; i <= n; i++) {
    if (i % block == 1 % block) ql[++tot] = i;
    qr[tot] = i, belong[i] = tot;
  }
  for (int i = 1; i <= tot; i++) res[i] = getres(ql[i], qr[i]);
  for (int i = 1; i <= q; i++) {
    int ql, qr, cur = 0;
    read(ql), read(qr);
    for (int j = 1; j <= tot; j++) cur = merge(cur, res[j].query(ql, qr));
    k[i] = cur;
  }
  printf("%d\n", size);
  for (auto x : ans) printf("%d %d\n", x.first, x.second);
  for (int i = 1; i <= q; i++) printf("%d ", k[i]);
  printf("\n");
  return 0;
}