Problem: Manao has invented a new mathematical term — a beautiful set of points. He calls a set of points on a plane beautiful if it meets the following conditions:

  1. The coordinates of each point in the set are integers. 
  2. For any two points from the set, the distance between them is a non-integer. 



Consider all points (x, y) which satisfy the inequations: 0 ≤ x ≤ n; 0 ≤ y ≤ m; x + y > 0. Choose their subset of maximum size such that it is also a beautiful set of points.

Input

The single line contains two space-separated integers n and m (1 ≤ n, m ≤ 100).

Output

In the first line print a single integer — the size k of the found beautiful set. In each of the next k lines print a pair of space-separated integers — the x- and y- coordinates, respectively, of a point from the set.

If there are several optimal solutions, you may print any of them.

Examples

Input

2 2


Output

3
0 1
1 2
2 0


Input

4 3


Output

4
0 3
2 1
3 0
4 2

Note

Consider the first sample. The distance between points (0, 1) and (1, 2) equals <image>, between (0, 1) and (2, 0) — <image>, between (1, 2) and (2, 0) — <image>. Thus, these points form a beautiful set. You cannot form a beautiful set with more than three points out of the given points. Note that this is not the only solution.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long int gcd(long long int a, long long int b) {
  if (a < b)
    return gcd(b, a);
  else if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
long long int isPrime(long long int n) {
  long long int p = (long long int)sqrt(n);
  for (long long int i = (2); i <= (p); i += (1))
    if (n % i == 0) return 0;
  return 1;
}
long long int pow(long long int b, long long int e) {
  if (e == 0)
    return 1;
  else if (e % 2 == 0) {
    long long int a = pow(b, e / 2);
    return a * a;
  } else {
    long long int a = pow(b, e / 2);
    return b * a * a;
  }
}
long long int powm(long long int x, long long int y,
                   long long int m = 1000000007) {
  x = x % m;
  long long int res = 1;
  while (y) {
    if (y & 1) res = res * x;
    res %= m;
    y = y >> 1;
    x = x * x;
    x %= m;
  }
  return res;
}
long long int lcm(long long int a, long long int b) {
  return (a * b) / gcd(a, b);
}
long long int modInverse(long long int a, long long int m) {
  return powm(a, m - 2, m);
}
bool issq(long long int n) {
  long long int p = sqrt(n);
  return p * p == n;
}
long long int MIN(long long int a = 2e18, long long int b = 2e18,
                  long long int c = 2e18, long long int d = 2e18,
                  long long int e = 2e18) {
  return min(a, min(b, min(c, min(d, e))));
}
long long int MAX(long long int a = -2e18, long long int b = -2e18,
                  long long int c = -2e18, long long int d = -2e18,
                  long long int e = -2e18) {
  return max(a, max(b, max(c, max(d, e))));
}
vector<long long int> prime;
void seive(long long int n) {
  prime.resize(n + 1);
  for (long long int i = (2); i <= (n); i += (1))
    if (prime[i] == 0) {
      prime[i] = i;
      for (long long int j = (i * i); j <= (n); j += (i))
        if (prime[j] == 0) prime[j] = i;
    }
}
long long int extended_GCD(long long int a, long long int b, long long int &x,
                           long long int &y) {
  if (a == 0) {
    x = 0;
    y = 1;
    return b;
  }
  long long int x1, y1;
  long long int gcd = extended_GCD(b % a, a, x1, y1);
  x = y1 - (b / a) * x1;
  y = x1;
  return gcd;
}
long long int mulInv(long long int a, long long int mod) {
  long long int x, y;
  extended_GCD(a, mod, x, y);
  if (x < 0) x += mod;
  return x;
}
template <typename T>
void swap(T *x, T *y) {
  T z = *x;
  *x = *y;
  *y = z;
}
template <typename T>
void print(T a[], long long int start, long long int end) {
  for (long long int i = (start); i <= (end); i += (1)) cout << a[i] << " ";
  cout << "\n";
}
template <typename T>
void print(T a, long long int start = 0, long long int end = 1e9) {
  end = min(end, (long long int)a.size() - 1);
  for (long long int i = (start); i <= (end); i += (1)) cout << a[i] << " ";
  cout << "\n";
}
template <typename T, typename P>
void print(vector<pair<T, P>> v, long long int start = 0,
           long long int end = 1e9) {
  end = min(end, (long long int)v.size() - 1);
  for (long long int i = (start); i <= (end); i += (1))
    cout << v[i].first << " " << v[i].second, cout << "\n";
  ;
}
long long int find(long long int num[], long long int rem[], long long int k,
                   long long int prod) {
  long long int result = 0;
  for (long long int i = 0; i < k; i++) {
    long long int pp = prod / num[i];
    result += rem[i] * mulInv(pp, num[i]) * pp;
  }
  return result % prod;
}
class DSU {
 public:
  vector<long long int> a, rank;

 public:
  long long int c;
  map<string, long long int> m;

 public:
  DSU(long long int n) {
    c = 1;
    a.resize(n + 1);
    rank.resize(n + 1);
    for (long long int i = 1; i <= n; i++) {
      a[i] = i;
      rank[i] = 1;
    }
  }

 public:
  long long int root(long long int x) {
    long long int u = x;
    while (a[x] != x) {
      a[x] = a[a[x]];
      x = a[x];
    }
    a[u] = x;
    return x;
  }

 public:
  void unite(long long int x, long long int y) {
    x = root(x);
    y = root(y);
    if (x != y) {
      if (rank[x] < rank[y])
        a[y] = x;
      else
        a[x] = y;
    }
  }
};
struct trip {
  long long int d, f, b, ind;
};
bool comp(pair<long long int, long long int> p1,
          pair<long long int, long long int> p2) {
  return (p1.first < p2.first ||
          (p1.first == p2.first && p1.second < p2.second));
}
vector<long long int> visi;
void dfs(long long int n, vector<long long int> adj[], long long int par) {
  long long int l = (adj[n]).size();
  visi[n] = 1;
  for (long long int i = (0); i <= (l - 1); i += (1)) {
    if (visi[(adj[n])[i]] == 0) {
      dfs((adj[n])[i], adj, n);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int n, m;
  cin >> n >> m;
  long long int x = MIN(n, m);
  cout << x + 1;
  cout << "\n";
  for (long long int i = (0); i <= (x); i += (1))
    cout << i << " " << x - i, cout << "\n";
  ;
  return 0;
}