#define _USE_MATH_DEFINES #include using namespace std; #define FOR(i,m,n) for(int i=(m);i<(n);++i) #define REP(i,n) FOR(i,0,n) #define ALL(v) (v).begin(),(v).end() using ll = long long; constexpr int INF = 0x3f3f3f3f; constexpr long long LINF = 0x3f3f3f3f3f3f3f3fLL; constexpr double EPS = 1e-8; constexpr int MOD = 1000000007; // constexpr int MOD = 998244353; constexpr int DY4[]{1, 0, -1, 0}, DX4[]{0, -1, 0, 1}; constexpr int DY8[]{1, 1, 0, -1, -1, -1, 0, 1}; constexpr int DX8[]{0, -1, -1, -1, 0, 1, 1, 1}; template inline bool chmax(T& a, U b) { return a < b ? (a = b, true) : false; } template inline bool chmin(T& a, U b) { return a > b ? (a = b, true) : false; } struct IOSetup { IOSetup() { std::cin.tie(nullptr); std::ios_base::sync_with_stdio(false); std::cout << fixed << setprecision(20); } } iosetup; template struct FenwickTree { explicit FenwickTree(const int n, const Abelian ID = 0) : n(n), ID(ID), data(n, ID) {} void add(int idx, const Abelian val) { for (; idx < n; idx |= idx + 1) { data[idx] += val; } } Abelian sum(int idx) const { Abelian res = ID; for (--idx; idx >= 0; idx = (idx & (idx + 1)) - 1) { res += data[idx]; } return res; } Abelian sum(const int left, const int right) const { return left < right ? sum(right) - sum(left) : ID; } Abelian operator[](const int idx) const { return sum(idx, idx + 1); } int lower_bound(Abelian val) const { if (val <= ID) return 0; int res = 0, exponent = 1; while (exponent <= n) exponent <<= 1; for (int mask = exponent >> 1; mask > 0; mask >>= 1) { const int idx = res + mask - 1; if (idx < n && data[idx] < val) { val -= data[idx]; res += mask; } } return res; } private: const int n; const Abelian ID; std::vector data; }; int main() { int n; cin >> n; vector p(n); REP(i, n) cin >> p[i], --p[i]; vector x(n), y(n), q(n); iota(ALL(x), 0); REP(i, n) q[p[i]] = i; FenwickTree bit(n); FOR(i, 1, n) { const int y1 = bit.sum(q[i - 1]), y2 = bit.sum(q[i]) + (q[i - 1] < q[i]); if (y1 % 2 == 1 || (x[i - 1] < x[i] && y2 % 2 == 1)) { swap(x[i - 1], x[i]); swap(q[i - 1], q[i]); } y[i - 1] = bit.sum(q[i - 1]); if (y[i - 1] % 2 == 1) { cout << "No\n"; return 0; } bit.add(q[i - 1], 1); } y[n - 1] = bit.sum(q[n - 1]); if (y[n - 1] % 2 == 1) { cout << "No\n"; } else { cout << "Yes\n"; REP(i, n) cout << x[i] + 1 << ' ' << y[i] + 1 << '\n'; } return 0; }