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>
const int MN = 1 << 12, MQ = 1 << 16;
const int S = 1 << 8, B = 1 << 4;
struct mrg {
 public:
  int a, b;
  void out() { printf("%d %d\n", a, b); }
};
std::vector<mrg> f;
std::vector<int> g;
int N, M, Q, cnt, a[MN], b[MN];
int get(int a, int b) {
  if (!~a) return b;
  if (!~b) return a;
  f.push_back({a, b});
  return ++cnt;
}
struct blk {
 public:
  int N, L, R;
  std::vector<int> v;
  std::vector<std::vector<int> > f;
  blk *l, *r;
  blk() {}
  blk(int _L, int _R) : L(_L), R(_R) {
    N = R - L;
    v.resize(N, -1);
    f.resize(N, {});
    std::iota(v.begin(), v.end(), L);
    assert(N > 0);
    std::sort(v.begin(), v.end(), [](int s, int t) { return b[s] < b[t]; });
    int M = L + (R - L) / 2;
    if (R - L > 1) {
      l = new blk(L, M);
      r = new blk(M, R);
    }
    for (int i = 0; i < N; ++i) {
      f[i].assign(N - i, -5);
      f[i][0] = b[v[i]];
      for (int j = 1; i + j < N; ++j)
        f[i][j] =
            get(l->ans(b[v[i]], b[v[i + j]]), r->ans(b[v[i]], b[v[i + j]]));
    }
  }
  int ans(int l, int r) {
    int i, j;
    for (i = 0; i < N && b[v[i]] < l; ++i)
      ;
    for (j = i; j < N && b[v[j]] <= r; ++j)
      ;
    if (i < j) return f[i][j - i - 1];
    return -1;
  }
} v[B];
int main() {
  scanf("%d%d", &N, &Q);
  cnt = N;
  for (int i = 0; i < N; ++i) scanf("%d", a + i), b[--a[i]] = i + 1;
  M = (N + S - 1) / S;
  assert(M <= B);
  for (int i = 0; i < M; ++i) v[i] = blk(i * S, std::min((i + 1) * S, N));
  for (int i = 0, l, r; i < Q; ++i) {
    int c = -1;
    scanf("%d%d", &l, &r);
    for (int j = 0; j < M; ++j) c = get(c, v[j].ans(l, r));
    g.push_back(c);
  }
  printf("%d\n", cnt);
  for (auto x : f) x.out();
  assert(g.size() == Q);
  for (int i = 0; i < Q; ++i) printf("%d%c", g[i], " \n"[i + 1 == Q]);
  return 0;
}