Former world number one Justine Henin will have the top seeds looking warily over their shoulders when she continues her fairytale comeback at the Indian Wells WTA tournament starting today. The Belgian has reached successive finals since returning to the circuit in January after quitting the game in 2008 and lurks as a dangerous wildcard entry at the desert venue where she won the women's title in 2004.

Henin, a seven-times grand slam champion, is scheduled to meet Slovakia's Magdalena Rybarikova in the first round with 31st-seeded Gisela Dulko of Argentina lying in wait as her next potential opponent. Should the 27-year-old Belgian advance to the third round, she is likely to face fifth seed Agneiszka Radwanska of Poland while fourth-seeded Russian Elena Dementieva is also in her quarter of the draw. Although Henin has not appeared on the circuit since losing to American Serena Williams in the Australian Open final in late January, she is still one of the best players of her generation and is hungry for further honours.

Tournament director Steve Simon is well aware of Henin's glowing stature and marketability, both on and off the court. Svetlana Kuznetsova is the top seed but the Russian faces a tricky path to the final, having been drawn to take on Belgian former world number one Kim Clijsters in the fourth round and sixth-seeded Serb Jelena Jankovic in the last eight.

Holder and 12th seed Vera Zvonareva of Russia is scheduled to come up against 2008 champion Ana Ivanovic of Serbia in a third-round rematch of last year's final.- Reuters