Problem: We call a string good, if after merging all the consecutive equal characters, the resulting string is palindrome. For example, "aabba" is good, because after the merging step it will become "aba".

Given a string, you have to find two values:

  1. the number of good substrings of even length; 
  2. the number of good substrings of odd length. 

Input

The first line of the input contains a single string of length n (1 ≤ n ≤ 105). Each character of the string will be either 'a' or 'b'.

Output

Print two space-separated integers: the number of good substrings of even length and the number of good substrings of odd length.

Examples

Input

bb


Output

1 2


Input

baab


Output

2 4


Input

babb


Output

2 5


Input

babaa


Output

2 7

Note

In example 1, there are three good substrings ("b", "b", and "bb"). One of them has even length and two of them have odd length.

In example 2, there are six good substrings (i.e. "b", "a", "a", "b", "aa", "baab"). Two of them have even length and four of them have odd length.

In example 3, there are seven good substrings (i.e. "b", "a", "b", "b", "bb", "bab", "babb"). Two of them have even length and five of them have odd length.

Definitions

A substring s[l, r] (1 ≤ l ≤ r ≤ n) of string s = s1s2... sn is string slsl + 1... sr.

A string s = s1s2... sn is a palindrome if it is equal to string snsn - 1... s1.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long Gcd(long long a, long long b) {
  if (a % b == 0) return b;
  return Gcd(b, a % b);
}
long long Max(long long a, long long b) {
  if (a > b) return a;
  return b;
}
long long Min(long long a, long long b) {
  if (a < b) return a;
  return b;
}
long long gcdExtended(long long a, long long b, long long *x, long long *y) {
  if (a == 0) {
    *x = 0, *y = 1;
    return b;
  }
  long long x1, y1;
  long long gcd = gcdExtended(b % a, a, &x1, &y1);
  *x = y1 - (b / a) * x1;
  *y = x1;
  return gcd;
}
long long power(long long x, long long y, long long p) {
  long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long modInverse(long long a, long long m) {
  long long x, y;
  long long g = gcdExtended(a, m, &x, &y);
  if (g != 1)
    return -1;
  else {
    long long res = (x % m + m) % m;
    return res;
  }
}
vector<int> primeFactors(int n) {
  vector<int> v;
  while (n % 2 == 0) {
    v.push_back(2);
    n = n / 2;
  }
  for (int i = 3; i <= sqrt(n); i = i + 2) {
    while (n % i == 0) {
      v.push_back(i);
      n = n / i;
    }
  }
  if (n > 2) v.push_back(n);
  return v;
}
long long C[2001][2000];
void binomialCoeff(long long n, long long k) {
  long long i, j;
  for (i = 0; i <= n; i++) {
    for (j = 0; j <= k; j++) {
      if (j == 0 || j == i)
        C[i][j] = 1;
      else
        C[i][j] = (C[i - 1][j - 1] % 1000000007 + C[i - 1][j] % 1000000007) %
                  1000000007;
    }
  }
}
long long spf[100010];
void sieve() {
  spf[1] = 1;
  for (int i = 2; i < 100010; i++) spf[i] = i;
  for (int i = 4; i < 100010; i += 2) spf[i] = 2;
  for (int i = 3; i * i < 100010; i++) {
    if (spf[i] == i) {
      for (int j = i * i; j < 100010; j += i)
        if (spf[j] == j) spf[j] = i;
    }
  }
}
long long binarySearch(long long arr[], long long l, long long r, long long x) {
  if (r >= l) {
    long long mid = l + (r - l) / 2;
    if (arr[mid] == x) return mid;
    if (arr[mid] > x) return binarySearch(arr, l, mid - 1, x);
    return binarySearch(arr, mid + 1, r, x);
  }
  return -1;
}
const long long N = 100000;
long long n;
long long tree[2 * N];
void build(long long arr[]) {
  for (long long i = 0; i < n; i++) {
    tree[n + i] = arr[i];
  }
  for (long long i = n - 1; i > 0; --i) {
    tree[i] = tree[i << 1] + tree[i << 1 | 1];
  }
}
void updateTreeNode(long long p, long long value) {
  tree[p + n] = value;
  p = p + n;
  for (long long i = p; i > 1; i >>= 1) {
    tree[i >> 1] = tree[i] + tree[i ^ 1];
  }
}
long long query(long long l, long long r) {
  long long res = 0;
  for (l += n, r += n; l < r; l >>= 1, r >>= 1) {
    if (l & 1) {
      res += tree[l++];
    }
    if (r & 1) {
      res += tree[--r];
    }
  }
  return res;
}
void dfs(std::vector<long long> v[], long long s, bool visited[]) {
  visited[s] = true;
  for (long long i = 0; i < v[s].size(); ++i) {
    if (!visited[v[s][i]]) {
      dfs(v, v[s][i], visited);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  string x69;
  cin >> x69;
  long long n = x69.length();
  char s[n + 1];
  for (int i = 0, _n = (n); i < _n; i++) {
    s[i + 1] = x69[i];
  }
  long long ev = 0;
  long long od = 0;
  long long ao = 0;
  long long ae = 0;
  long long bo = 0;
  long long be = 0;
  for (int i = (1), _b = (n); i <= _b; i++) {
    if (s[i] == 'a') {
      if (i % 2) {
        od += ao + 1;
        ev += ae;
      } else {
        od += ae + 1;
        ev += ao;
      }
      if (i % 2)
        ao++;
      else
        ae++;
    }
    if (s[i] == 'b') {
      if (i % 2) {
        od += bo + 1;
        ev += be;
      } else {
        od += be + 1;
        ev += bo;
      }
      if (i % 2)
        bo++;
      else
        be++;
    }
  }
  cout << ev << " " << od << endl;
  return 0;
}