Problem: There are two strings s and t, consisting only of letters a and b. You can make the following operation several times: choose a prefix of s, a prefix of t and swap them. Prefixes can be empty, also a prefix can coincide with a whole string. 

Your task is to find a sequence of operations after which one of the strings consists only of a letters and the other consists only of b letters. The number of operations should be minimized.

Input

The first line contains a string s (1 ≤ |s| ≤ 2·105).

The second line contains a string t (1 ≤ |t| ≤ 2·105).

Here |s| and |t| denote the lengths of s and t, respectively. It is guaranteed that at least one of the strings contains at least one a letter and at least one of the strings contains at least one b letter.

Output

The first line should contain a single integer n (0 ≤ n ≤ 5·105) — the number of operations.

Each of the next n lines should contain two space-separated integers ai, bi — the lengths of prefixes of s and t to swap, respectively.

If there are multiple possible solutions, you can print any of them. It's guaranteed that a solution with given constraints exists.

Examples

Input

bab
bb


Output

2
1 0
1 3


Input

bbbb
aaa


Output

0

Note

In the first example, you can solve the problem in two operations:

  1. Swap the prefix of the first string with length 1 and the prefix of the second string with length 0. After this swap, you'll have strings ab and bbb. 
  2. Swap the prefix of the first string with length 1 and the prefix of the second string with length 3. After this swap, you'll have strings bbbb and a. 



In the second example, the strings are already appropriate, so no operations are needed.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <typename T, typename U>
std::istream& operator>>(std::istream& i, pair<T, U>& p) {
  i >> p.first >> p.second;
  return i;
}
template <typename T>
std::istream& operator>>(std::istream& i, vector<T>& t) {
  for (auto& v : t) {
    i >> v;
  }
  return i;
}
template <typename T, typename U>
std::ostream& operator<<(std::ostream& o, const pair<T, U>& p) {
  o << p.first << ' ' << p.second;
  return o;
}
template <typename T>
std::ostream& operator<<(std::ostream& o, const vector<T>& t) {
  if (t.empty()) o << '\n';
  for (size_t i = 0; i < t.size(); ++i) {
    o << t[i] << " \n"[i == t.size() - 1];
  }
  return o;
}
template <typename T>
using minheap = priority_queue<T, vector<T>, greater<T>>;
template <typename T>
using maxheap = priority_queue<T, vector<T>, less<T>>;
template <typename T>
bool in(T a, T b, T c) {
  return a <= b && b < c;
}
unsigned int logceil(int first) {
  return 8 * sizeof(int) - __builtin_clz(first);
}
namespace std {
template <typename T, typename U>
struct hash<pair<T, U>> {
  hash<T> t;
  hash<U> u;
  size_t operator()(const pair<T, U>& p) const {
    return t(p.first) ^ (u(p.second) << 7);
  }
};
}  // namespace std
template <typename T, typename F>
T bsh(T l, T h, const F& f) {
  T r = -1, m;
  while (l <= h) {
    m = (l + h) / 2;
    if (f(m)) {
      l = m + 1;
      r = m;
    } else {
      h = m - 1;
    }
  }
  return r;
}
template <typename F>
double bshd(double l, double h, const F& f, double p = 1e-9) {
  unsigned int r = 3 + (unsigned int)log2((h - l) / p);
  while (r--) {
    double m = (l + h) / 2;
    if (f(m)) {
      l = m;
    } else {
      h = m;
    }
  }
  return (l + h) / 2;
}
template <typename T, typename F>
T bsl(T l, T h, const F& f) {
  T r = -1, m;
  while (l <= h) {
    m = (l + h) / 2;
    if (f(m)) {
      h = m - 1;
      r = m;
    } else {
      l = m + 1;
    }
  }
  return r;
}
template <typename F>
double bsld(double l, double h, const F& f, double p = 1e-9) {
  unsigned int r = 3 + (unsigned int)log2((h - l) / p);
  while (r--) {
    double m = (l + h) / 2;
    if (f(m)) {
      h = m;
    } else {
      l = m;
    }
  }
  return (l + h) / 2;
}
template <typename T>
T gcd(T a, T b) {
  if (a < b) swap(a, b);
  return b ? gcd(b, a % b) : a;
}
template <typename T>
class vector2 : public vector<vector<T>> {
 public:
  vector2() {}
  vector2(size_t a, size_t b, T t = T())
      : vector<vector<T>>(a, vector<T>(b, t)) {}
};
template <typename T>
class vector3 : public vector<vector2<T>> {
 public:
  vector3() {}
  vector3(size_t a, size_t b, size_t c, T t = T())
      : vector<vector2<T>>(a, vector2<T>(b, c, t)) {}
};
template <typename T>
class vector4 : public vector<vector3<T>> {
 public:
  vector4() {}
  vector4(size_t a, size_t b, size_t c, size_t d, T t = T())
      : vector<vector3<T>>(a, vector3<T>(b, c, d, t)) {}
};
template <typename T>
class vector5 : public vector<vector4<T>> {
 public:
  vector5() {}
  vector5(size_t a, size_t b, size_t c, size_t d, size_t e, T t = T())
      : vector<vector4<T>>(a, vector4<T>(b, c, d, e, t)) {}
};
class TaskD {
 public:
  void solve(istream& cin, ostream& cout) {
    string S, T;
    cin >> S >> T;
    auto runs = [](const string& S) {
      int i = 0, r = 0;
      while (i < S.size()) {
        char z = S[i];
        while (i < S.size() && S[i] == z) ++i;
        ++r;
      }
      return r;
    };
    vector<std::pair<int, int>> Ans;
    int best = 1e9;
    auto adv = [](int& i, int& s, const string& S) {
      if (i == S.size()) return;
      char z = S[i];
      while (i < S.size() && S[i] == z) {
        ++i;
        ++s;
      }
    };
    auto advRuns = [&](string& S, int runs) {
      int i = 0;
      while (i < S.size() && runs > 0) {
        char z = S[i];
        while (i < S.size() && S[i] == z) ++i;
        --runs;
      }
      return i;
    };
    auto solve = [&](int r1, int r2, int w) {
      vector<std::pair<int, int>> Cur;
      int s = 0, t = 0;
      int i = advRuns(S, r1);
      int j = advRuns(T, r2);
      if (i != 0 || j != 0) {
        Cur.push_back({i, j});
      }
      string A = T.substr(0, j) + S.substr(i);
      string B = S.substr(0, i) + T.substr(j);
      i = 0, j = 0;
      if (A[0] == B[0]) {
        if (w) {
          adv(i, s, A);
        } else {
          adv(j, t, B);
        }
        Cur.push_back({i, j});
        swap(s, t);
      }
      while (i < A.size() || j < B.size()) {
        adv(i, s, A);
        adv(j, t, B);
        if (i >= A.size() && j >= B.size()) break;
        Cur.push_back({s, t});
        swap(t, s);
      }
      if (Cur.size() < best) {
        Ans = Cur;
        best = Cur.size();
      }
    };
    int r1 = runs(S);
    int r2 = runs(T);
    int q = 4;
    for (int w = 0; w < 2; ++w) {
      for (int z = max(0, (r1 - r2) / 2 - q); z <= min(r1, (r1 - r2) / 2 + q);
           ++z) {
        solve(z, 0, w);
        solve(z, 1, w);
      }
      for (int z = max(0, (r2 - r1) / 2 - q); z <= min(r2, (r2 - r1) / 2 + q);
           ++z) {
        solve(0, z, w);
        solve(1, z, w);
      }
    }
    cout << Ans.size() << '\n';
    for (auto ans : Ans) cout << ans << '\n';
  }
};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  TaskD solver;
  std::istream& in(std::cin);
  std::ostream& out(std::cout);
  solver.solve(in, out);
  return 0;
}