Problem: East or west, home is best. That's why family reunion, the indispensable necessity of Lunar New Year celebration, is put in such a position.

After the reunion dinner, Little Tommy plays a game with the family. Here is a concise introduction to this game: 

  1. There is a sequence of n non-negative integers p1, p2, ..., pn in the beginning. It is ruled that each integer in this sequence should be non-negative at any time. 
  2. You can select two consecutive positive integers in this sequence, pi and pi + 1 (1 ≤ i < n), and then decrease them by their minimum (i. e. min(pi, pi + 1)), the cost of this operation is equal to min(pi, pi + 1). We call such operation as a descension. 
  3. The game immediately ends when there are no two consecutive positive integers. Your task is to end the game so that the total cost of your operations is as small as possible. 



Obviously, every game ends after at most n - 1 descensions. Please share your solution of this game with the lowest cost.

Input

The first line contains one integer n (1 ≤ n ≤ 3·105).

The second line contains n space-separated integers p1, p2, ..., pn (0 ≤ pi ≤ 109, i = 1, 2, ..., n).

Output

In the first line print one integer as the number of descensions m (0 ≤ m ≤ n - 1).

In the next m lines print the descensions chronologically. More precisely, in each line of the next m lines print one integer i (1 ≤ i < n) representing a descension would operate on pi and pi + 1 such that all the descensions could be utilized from top to bottom.

If there are many possible solutions to reach the minimal cost, print any of them.

Examples

Input

4
2 1 3 1


Output

2
1
3


Input

5
2 2 1 3 1


Output

3
2
1
4

Note

In the first sample, one possible best solution is <image>, of which the cost is 1 + 1 = 2.

In the second sample, one possible best solution is <image>, of which the cost is 1 + 1 + 1 = 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
bool debug = 0;
int n, m, k;
int dx[4] = {0, 1, 0, -1}, dy[4] = {1, 0, -1, 0};
string direc = "RDLU";
long long ln, lk, lm;
void etp(bool f = 0) {
  puts(f ? "YES" : "NO");
  exit(0);
}
void addmod(int &x, int y, int mod = 1000000007) {
  assert(y >= 0);
  x += y;
  if (x >= mod) x -= mod;
  assert(x >= 0 && x < mod);
}
void et() {
  puts("-1");
  exit(0);
}
long long fastPow(long long x, long long y, int mod = 1000000007) {
  long long ans = 1;
  while (y > 0) {
    if (y & 1) ans = (x * ans) % mod;
    x = x * x % mod;
    y >>= 1;
  }
  return ans;
}
long long gcd1(long long x, long long y) {
  long long z = y;
  while (x % y != 0) {
    z = x % y;
    x = y;
    y = z;
  }
  return z;
}
int p[300105];
long long dp[300105];
int pre[300105];
vector<int> ans;
void wk(int i) {
  if (i <= 0) return;
  if (p[i] == 0 || p[i + 1] == 0) return;
  ans.push_back(i);
  int t = min(p[i], p[i + 1]);
  p[i] -= t;
  p[i + 1] -= t;
}
void fmain(int ID) {
  scanf("%d", &n);
  for (int(i) = 1; (i) <= (int)(n); (i)++) scanf("%d", p + i);
  for (int(i) = 1; (i) <= (int)(n); (i)++) {
    if (dp[max(i - 2, 0)] + p[i] < dp[max(i - 3, 0)] + max(p[i], p[i - 1])) {
      dp[i] = dp[max(i - 2, 0)] + p[i];
      pre[i] = max(i - 2, 0);
    } else {
      dp[i] = dp[max(i - 3, 0)] + max(p[i], p[i - 1]);
      pre[i] = max(i - 3, 0);
    }
  }
  long long A = min(dp[n - 1], dp[n]);
  for (int i = dp[n - 1] < dp[n] ? n - 1 : n; i; i = pre[i]) {
    if (pre[i] == max(i - 3, 0)) {
      wk(i - 1);
    }
    wk(pre[i] + 1);
    wk(i);
  }
  printf("%d\n", (int)ans.size());
  for (int c : ans) printf("%d\n", c);
}
int main() {
  int t = 1;
  for (int(i) = 1; (i) <= (int)(t); (i)++) {
    fmain(i);
  }
  return 0;
}