Problem: A group of tourists is going to kayak and catamaran tour. A rented lorry has arrived to the boat depot to take kayaks and catamarans to the point of departure. It's known that all kayaks are of the same size (and each of them occupies the space of 1 cubic metre), and all catamarans are of the same size, but two times bigger than kayaks (and occupy the space of 2 cubic metres).

Each waterborne vehicle has a particular carrying capacity, and it should be noted that waterborne vehicles that look the same can have different carrying capacities. Knowing the truck body volume and the list of waterborne vehicles in the boat depot (for each one its type and carrying capacity are known), find out such set of vehicles that can be taken in the lorry, and that has the maximum total carrying capacity. The truck body volume of the lorry can be used effectively, that is to say you can always put into the lorry a waterborne vehicle that occupies the space not exceeding the free space left in the truck body.

Input

The first line contains a pair of integer numbers n and v (1 ≤ n ≤ 105; 1 ≤ v ≤ 109), where n is the number of waterborne vehicles in the boat depot, and v is the truck body volume of the lorry in cubic metres. The following n lines contain the information about the waterborne vehicles, that is a pair of numbers ti, pi (1 ≤ ti ≤ 2; 1 ≤ pi ≤ 104), where ti is the vehicle type (1 – a kayak, 2 – a catamaran), and pi is its carrying capacity. The waterborne vehicles are enumerated in order of their appearance in the input file.

Output

In the first line print the maximum possible carrying capacity of the set. In the second line print a string consisting of the numbers of the vehicles that make the optimal set. If the answer is not unique, print any of them.

Examples

Input

3 2
1 2
2 7
1 3


Output

7
2

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
mt19937_64 rang(
    chrono::high_resolution_clock::now().time_since_epoch().count());
long long mpow(long long base, long long exp);
long long power(long long a, long long b);
const long long mod = 1000000007;
vector<int> g[300001];
struct node {
  set<int> s;
};
struct segTree {
  long long arr[2001];
  long long res[8003];
  int idx_mx[8003];
  int operation_arr[4003];
  int NO_OPERATION = 0;
  void init(int x, int lo, int hi) {
    if (lo == hi) {
      res[x] = arr[lo];
      return;
    }
    int mid = (lo + hi) / 2;
    init(2 * x, lo, mid);
    init(2 * x + 1, mid + 1, hi);
    update_point(x);
  }
  int merge_operation(int a, int b) {
    if (b != NO_OPERATION) return (b);
    return a;
  }
  long long merge_results(int a, int b) { return a + b; }
  void modify(int x, int lo, int hi, int l, int r, int op) {
    prop(x, lo, hi);
    if (lo >= l && hi <= r) {
      operation_arr[x] = merge_operation(operation_arr[x], op);
      return;
    }
    int mid = (lo + hi) / 2;
    if (lo > r || hi < l) return;
    modify(2 * x, lo, mid, l, r, op);
    modify(2 * x + 1, mid + 1, hi, l, r, op);
    update(x, lo, hi);
  }
  long long get_sum(int x, int lo, int hi, int l, int r) {
    if (lo >= l && hi <= r) return res[x];
    if (lo > r || hi < l) return 0;
    int mid = (lo + hi) / 2;
    return get_sum(2 * x, lo, mid, l, r) +
           get_sum(2 * x + 1, mid + 1, hi, l, r);
  }
  void point_modify(int x, int lo, int hi, int pos, long long val) {
    if (lo == hi) {
      res[x] = val;
      return;
    }
    int mid = (lo + hi) / 2;
    if (pos <= mid)
      point_modify(2 * x, lo, mid, pos, val);
    else
      point_modify(2 * x + 1, mid + 1, hi, pos, val);
    update_point(x);
  }
  long long max_query(int x, int lo, int hi, int l, int r) {
    if (lo >= l && hi <= r) return res[x];
    if (lo > r || hi < l) return LLONG_MIN;
    int mid = (lo + hi) / 2;
    return max(max_query(2 * x, lo, mid, l, r),
               max_query(2 * x + 1, mid + 1, hi, l, r));
  }
  long long get_idx(int x, int lo, int hi, long long v) {
    if (res[x] < v) return hi + 1;
    if (lo == hi) return lo;
    int mid = (lo + hi) / 2;
    if (res[2 * x] >= v)
      return get_idx(2 * x, lo, mid, v);
    else
      return get_idx(2 * x + 1, mid + 1, hi, v);
  }
  pair<int, int> mx_q(int x, int lo, int hi, int l, int r) {
    if (lo >= l && hi <= r) return {res[x], idx_mx[x]};
    if (lo > r || hi < l) return {LLONG_MIN, -1};
    int mid = (lo + hi) / 2;
    pair<int, int> p1 = mx_q(2 * x, lo, mid, l, r);
    pair<int, int> p2 = mx_q(2 * x + 1, mid + 1, hi, l, r);
    if (p1.first > p2.first) {
      return p1;
    } else
      return p2;
  }
  int mx_idx = -1;
  void max_idx(int x, int lo, int hi, int val, int r) {
    if (res[x] < val) return;
    int mid = (lo + hi) / 2;
    if (lo > r) return;
    if (hi == lo && res[x] == val) {
      mx_idx = lo;
      return;
    }
    max_idx(2 * x, lo, mid, val, r);
    max_idx(2 * x + 1, mid + 1, hi, val, r);
  }
  int get_max_idx(int x, int lo, int hi, int val, int r) {
    max_idx(x, lo, hi, val, r);
    return mx_idx;
  }
  void update_point(int x) { res[x] = res[2 * x] + res[2 * x + 1]; }
  int query(int x, int lo, int hi, int l, int r) {
    prop(x, lo, hi);
    if (lo >= l && hi <= r) return operate(res[x], operation_arr[x], lo, hi);
    if (lo > r || hi < l) return 0;
    int mid = (lo + hi) / 2;
    return merge_results(query(2 * x, lo, mid, l, r),
                         query(2 * x + 1, mid + 1, hi, l, r));
  }
  int kth_one(int x, int lo, int hi, int k) {
    if (lo == hi) return lo;
    int v1 = res[2 * x];
    int mid = (lo + hi) / 2;
    if (k <= v1)
      return kth_one(2 * x, lo, mid, k);
    else
      return kth_one(2 * x + 1, mid + 1, hi, k - v1);
  }
  long long point_query(int x, int lo, int hi, int p) {
    prop(x, lo, hi);
    if (lo == hi) return operate(res[x], operation_arr[x], lo, lo);
    int mid = (lo + hi) / 2;
    if (p <= mid)
      return point_query(2 * x, lo, mid, p);
    else
      return point_query(2 * x + 1, mid + 1, hi, p);
  }
  void prop(int x, int lo, int hi) {
    if (lo == hi) return;
    int mid = (hi + lo) / 2;
    long long len = mid - lo + 1;
    operation_arr[2 * x] =
        merge_operation(operation_arr[2 * x], operation_arr[x]);
    operation_arr[2 * x + 1] =
        merge_operation(operation_arr[2 * x + 1], operation_arr[x]);
    operation_arr[x] = NO_OPERATION;
    update(x, lo, hi);
  }
  int operate(int val, int operatr, int l, int r) {
    if (operatr == NO_OPERATION) return NO_OPERATION;
    return operatr;
  }
  void update(int x, int lo, int hi) {
    if (lo == hi) return;
    int mid = (lo + hi) / 2;
    long long len1 = (mid - lo + 1);
    long long len2 = (hi - mid);
    res[x] = merge_results(
        operate(res[2 * x], operation_arr[2 * x], lo, mid),
        operate(res[2 * x + 1], operation_arr[2 * x + 1], mid + 1, hi));
  }
};
struct FenwickTree {
  int arr[100];
  int n;
  void update(int i, int delta) {
    while (i <= n) {
      arr[i] += delta;
      i = i + (i & (-i));
    }
  }
  int sum(int i) {
    if (i == 0) return 0;
    int res = 0;
    while (i > 0) {
      res = res + arr[i];
      i -= (i & (-i));
    }
    return res;
  }
};
struct dsu {
  int par[1001];
  int size_set[1001];
  void init(int n) {
    for (int i = 0; i <= n; i++) par[i] = i, size_set[i] = 1;
  }
  int get_par(int u) {
    vector<int> temp;
    while (par[u] != u) {
      temp.push_back(u);
      u = par[u];
    }
    for (int i : temp) par[i] = u;
    return u;
  }
  void union_set(int a, int b) {
    a = get_par(a);
    b = get_par(b);
    if (a == b) return;
    if (size_set[a] < size_set[b]) {
      int temp = a;
      a = b;
      b = temp;
    }
    par[b] = a;
    size_set[a] += size_set[b];
  }
  int get_size(int u) {
    u = get_par(u);
    return size_set[u];
  }
};
int check(string s1, string s2) {
  int n1 = s1.size();
  int n2 = s2.size();
  s1 = '$' + s1;
  s2 = '$' + s2;
  int dp[n1 + 1][n2 + 1];
  for (int j = 0; j <= n2; j++) dp[0][j] = 0;
  dp[0][0] = 1;
  for (int i = 1; i <= n1; i++) {
    dp[i][0] = 0;
  }
  int i1 = 1;
  while (i1 <= n1 && s1[i1] == '*') {
    dp[i1][0] = 1;
    i1++;
  }
  int sum[n2 + 1];
  int s = 0;
  for (int i = 0; i <= n2; i++) s = s + dp[0][i], sum[i] = s;
  for (int i = 1; i <= n1; i++) {
    for (int j = 1; j <= n2; j++) {
      char c1 = s1[i];
      char c2 = s2[j];
      if (c1 == '*') {
        if (sum[j] > 0)
          dp[i][j] = 1;
        else
          dp[i][j] = 0;
      } else if (c1 == '?') {
        dp[i][j] = dp[i - 1][j - 1];
      } else {
        if (c1 != c2)
          dp[i][j] = 0;
        else
          dp[i][j] = dp[i - 1][j - 1];
      }
    }
    s = 0;
    for (int j = 0; j <= n2; j++) {
      s += dp[i][j];
      sum[j] = s;
    }
  }
  return dp[n1][n2];
}
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  if (b == 0) return a;
  return gcd(b, a % b);
}
int a[100002];
int curr_ans = 0;
map<int, int> coded;
int code2[100001];
bool contains_dup(string s) {
  int cnt[26];
  memset(cnt, 0, sizeof(cnt));
  for (int i = 0; i < s.size(); i++) {
    cnt[s[i] - 'a']++;
    if (cnt[s[i] - 'a'] > 1) return 1;
  }
  return 0;
}
bool isOverflow(long long a, long long b) {
  if (a == 0 || b == 0) return false;
  if (a <= LLONG_MAX / b)
    return false;
  else
    return true;
}
bool isOverflow_add(long long a, long long b) {
  long long res = a + b;
  if (res - a == b && res - b == a) return 0;
  return 1;
}
long long fact(long long i) {
  long long res = 1;
  for (int j = 1; j <= i; j++) res *= j;
  return res;
}
long long f1[5001];
long long f2[5001];
void fct() {
  f1[0] = 1;
  f2[0] = mpow(1, mod - 2);
  for (long long i = 1; i <= 5000; i++) {
    f1[i] = (f1[i - 1] * i) % mod;
    f2[i] = mpow(f1[i], mod - 2);
  }
}
long long ncr(int n, int r) {
  long long res = f1[n];
  res = (res * f2[r]) % mod;
  res = (res * f2[n - r]) % mod;
  return res;
}
void solve() {
  int n, v;
  cin >> n >> v;
  vector<pair<long long, long long> > c, k;
  for (int i = 0; i < n; i++) {
    long long t, w;
    cin >> t >> w;
    if (t == 1)
      k.push_back({w, i});
    else
      c.push_back({w, i});
  }
  sort(c.begin(), c.end());
  sort(k.begin(), k.end());
  long long n1 = (int)c.size();
  long long n2 = (int)k.size();
  long long suf_sum1[n1 + 1];
  long long suf_sum2[n2 + 1];
  suf_sum1[n1] = 0;
  suf_sum2[n2] = 0;
  for (int i = n2 - 1; i >= 0; i--) suf_sum2[i] = suf_sum2[i + 1] + k[i].first;
  for (int i = n1 - 1; i >= 0; i--) suf_sum1[i] = suf_sum1[i + 1] + c[i].first;
  int cc = 0, ck = 0;
  long long res = 0;
  for (int cat = 0; cat <= n1; cat++) {
    long long vol_cat = cat * 2;
    long long vol_rem = v - vol_cat;
    long long kayak = min(n2, vol_rem);
    if (kayak < 0) break;
    long long sum = suf_sum1[n1 - cat] + suf_sum2[n2 - kayak];
    if (sum > res) {
      res = sum;
      cc = cat;
      ck = kayak;
    }
  }
  cout << res << "\n";
  vector<int> res_p;
  for (int i = n1 - cc; i < n1; i++) res_p.push_back(c[i].second);
  for (int i = n2 - ck; i < n2; i++) res_p.push_back(k[i].second);
  sort(res_p.begin(), res_p.end());
  for (int i : res_p) cout << i + 1 << " ";
  cout << "\n";
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  srand(chrono::high_resolution_clock::now().time_since_epoch().count());
  int t = 1;
  int t1 = 1;
  while (t--) {
    solve();
    t1++;
  }
  return 0;
}
long long mpow(long long base, long long exp) {
  base %= mod;
  long long result = 1;
  while (exp > 0) {
    if (exp & 1) result = ((long long)result * base) % mod;
    base = ((long long)base * base) % mod;
    exp >>= 1;
  }
  return result;
}
long long power(long long a, long long b) {
  long long res = 1;
  while (b > 0) {
    if (b % 2 == 0) {
      b /= 2;
      if (isOverflow(a, a)) return LLONG_MAX;
      a = a * a;
    } else {
      if (isOverflow(res, a)) return LLONG_MAX;
      res = res * a;
      b--;
    }
  }
  return res;
}